Limit, continuity and derivative

Class 12 Mathematics, unit 2, book pages 43–88

This unit builds calculus on one idea, the limit. It shows how to find limits and test continuity, defines the derivative as the limit of a difference quotient, and develops rules for differentiating algebraic, trigonometric, inverse trigonometric, exponential, logarithmic, implicit and parametric functions. The derivative is then used to approximate values and errors, to find where a function increases or decreases, to locate maxima and minima, and to solve rate and optimisation problems.

Learning outcomes (16)
  • Demonstrate and find the limit of a function.
  • State and apply theorems on the limit of a sum, difference, product and quotient of functions to algebraic, exponential and trigonometric functions.
  • Demonstrate and test continuity and discontinuity of a function at a point and in an interval.
  • Apply concepts of transcendental functions, the limit of a function and its continuity to real-world problems.
  • Calculate inflation over a period. Calculate depreciation with the help of the straight-line method.
  • Recognize the meaning of the tangent to a curve at a point.
  • Calculate the gradient of a curve at a point. Identify the derivative as the limit of a difference quotient. Calculate the derivative of a function. Estimate the derivative as a rate of change of velocity, temperature and profit. Recognize the derivative function.
  • State the connection between derivative and continuity.
  • Find the derivative of functions, including square root, quadratic and logarithmic functions.
  • Apply the differentiation rules to polynomials, rational and trigonometric functions.
  • Apply differentiation to state whether a function is increasing or decreasing.
  • Apply differentiation to real-world problems.
  • Find higher order derivatives of algebraic, implicit, parametric, trigonometric and inverse trigonometric functions. Describe the ability to approximate functions.
  • Explain differentials to approximate the change in a quantity. Calculate errors.
  • Find extreme values by applying the second derivative test. Explain and find critical points.
  • Apply the derivative and higher order derivatives to real-world problems.

2.1Limit of a functionOften examined

Calculus is the mathematics of change. Its first tool is the limit. The limit of as approaches a number is the value that gets closer and closer to when gets closer and closer to . We write as , and read the arrow as "approaches".

A first example

Take . Its domain is every real number except , because makes the denominator zero, so is not defined. We can still work out for values of near .

As gets close to from either side, gets close to .
(from the left) (from the right)

The table suggests that approaches . Algebra confirms it. For the factor cancels:

−7
−6
−5
−4
−3
−2
−1
1
2
3
2
4
6
8
10
12
O
The graph of is the line with a hole at . The limit is even though does not exist.
Limit (intuitive definition)
If can be made as close as we like to a finite number by taking close enough to (but not equal to ), from both the left and the right, then is the limit of as . We write .
Left-hand limit
: the value approaches when approaches from the left, that is, through values less than .
Right-hand limit
: the value approaches when approaches from the right, that is, through values greater than .

The limit exists only when both one-sided limits exist and are equal.

Example 1 (book): limit of a piecewise function
Using the graph, check whether exists, where
  1. For the rule is , so approach from the left using it. The table , , agrees.
  2. For the rule is . The table , , agrees.
  3. The two one-sided limits are different, so the graph has a jump at .
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
O
The left piece ends at a filled dot ; the right piece starts at a hollow dot .

Why a limit may fail to exist

  • Jump: the left and right limits are different numbers (Example 1, and at ).
  • Unbounded values: grows without bound near , for example near . We may write , but is not a number, so the limit does not exist.
  • Oscillation: keeps jumping between values, for example near takes every value from to in every small interval around .

2.1Theorems on limits and methods of evaluationOften examined

Tables and graphs show what a limit is, but they are slow. The theorems below let us find limits by algebra. Suppose and both exist, is a constant and is a positive integer.

Theorems on limits. The book's box lists the constant, constant multiple, sum, quotient and power theorems; the product and root theorems are used in the same way.
TheoremStatement
Constant and
Constant multiple
Sum and difference
Product
Quotient, provided
Power
Root (for even we need )

Direct substitution. Applying these theorems term by term shows that for any polynomial , . For a rational function with , the limit is . So the first thing to try is always to put .

A method that always works in the exam

  1. Substitute . If the answer is a real number (and no denominator is zero), that number is the limit.
  2. If you get , the form is indeterminate: it tells you nothing yet. Simplify first: factorise and cancel the common factor, or rationalise a surd, or use a standard limit such as . Then substitute again.
  3. If you get with , the function grows without bound near , so the limit does not exist as a finite number. Check the sign from each side if you need to say or .
Example 3 (book): direct substitution and factorising
Evaluate (a) (b) (c) (d) (e) (f)
  1. (a) The limit of a constant is the constant.
  2. (b) Constant multiple theorem.
  3. (c) A polynomial: substitute directly.
  4. (d) The denominator is , so the quotient theorem applies.
  5. (e) Substitution gives because . Factorise the denominator and cancel (allowed because while ).
  6. (f) Power theorem.
Example 4 (book): when the limit does not exist, and a cube root
Evaluate (a) (b)
  1. (a) The numerator tends to and the denominator to . The form is , not , so no cancelling can remove the zero. The values of the fraction grow without bound near .
  2. (b) The denominator tends to , so the quotient theorem applies. Note .
Example: rationalising a surd
Evaluate .
  1. Substitution gives . Multiply top and bottom by the conjugate to remove the surd from the numerator.
  2. Now substitute .

2.1Important limitsOften examined

Some limits have the form \dfrac00 but cannot be simplified by factorising. The two trigonometric ones below are the foundation for the derivatives of \sin x and \cos x (section 2.9). In all of them the angle x is in radians.

Example 2 (book): the two trigonometric limits

Values near 0 (the sign \pm follows the sign of x). The first ratio is even, the second is odd.
x\dfrac{\sin x}{x}\dfrac{1-\cos x}{x}
\pm0.10.998334\pm0.0499583
\pm0.010.999983\pm0.0049999583
\pm0.0010.9999998\pm0.0005000
\pm0.00010.9999999983\pm0.0000500
\lim_{x\to0}\dfrac{\sin x}{x}=1\qquad\qquad\lim_{x\to0}\dfrac{1-\cos x}{x}=0

Derivation Proof that \lim_{\theta\to0}\dfrac{\sin\theta}{\theta}=1

Aim. Show that \dfrac{\sin\theta}{\theta}\to1 as \theta\to0, using areas in a unit circle.

Assumptions.
  • \theta is measured in radians, so the area of a sector of radius r is \tfrac12r^{2}\theta.
  • First take 0<\theta<\tfrac{\pi}{2}; the case \theta<0 is handled at the end.
Unit circle, centre O. Triangle OAP lies inside sector OAP, which lies inside triangle OAT (tangent at A).
  1. Draw a unit circle (radius 1) with centre O. Let A be (1,0), P the point (\cos\theta,\sin\theta) on the circle, and T the point where OP extended meets the tangent at A. Then AT=\tan\theta and the height of P above OA is PM=\sin\theta.
  2. Compare three areas. Triangle OAP lies inside sector OAP, which lies inside triangle OAT.
    \tfrac12(1)(\sin\theta)<\tfrac12(1)^{2}\theta<\tfrac12(1)(\tan\theta)
  3. Divide every part by \tfrac12\sin\theta, which is positive, so the inequality signs stay the same.
    1<\dfrac{\theta}{\sin\theta}<\dfrac{1}{\cos\theta}
  4. Take reciprocals of positive numbers; this reverses the inequalities.
    \cos\theta<\dfrac{\sin\theta}{\theta}<1
  5. As \theta\to0^{+}, \cos\theta\to1. The ratio is trapped between two quantities that both tend to 1 (sandwich theorem), so it also tends to 1.
    \lim_{\theta\to0^{+}}\dfrac{\sin\theta}{\theta}=1
  6. For \theta<0 put \theta=-\phi with \phi>0. Since \sin(-\phi)=-\sin\phi, the ratio is unchanged, so the left-hand limit is also 1.
    \dfrac{\sin(-\phi)}{-\phi}=\dfrac{\sin\phi}{\phi}\to1
Result\lim_{\theta\to0}\dfrac{\sin\theta}{\theta}=1

For small angles in radians, \sin\theta\approx\theta. This result is used to find \dfrac{d}{dx}\sin x.

Derivation Proof that \lim_{x\to0}\dfrac{1-\cos x}{x}=0

Aim. Reduce the limit to \dfrac{\sin x}{x}.

  1. Multiply the top and bottom by 1+\cos x so that the numerator becomes 1-\cos^{2}x.
    \dfrac{1-\cos x}{x}=\dfrac{(1-\cos x)(1+\cos x)}{x(1+\cos x)}=\dfrac{1-\cos^{2}x}{x(1+\cos x)}
  2. Use 1-\cos^{2}x=\sin^{2}x and split into two factors whose limits we know.
    =\dfrac{\sin x}{x}\cdot\dfrac{\sin x}{1+\cos x}
  3. Take limits using the product theorem: \dfrac{\sin x}{x}\to1, \sin x\to0 and 1+\cos x\to2.
    \lim_{x\to0}\dfrac{1-\cos x}{x}=1\cdot\dfrac{0}{2}=0
Result\lim_{x\to0}\dfrac{1-\cos x}{x}=0

Near 0, 1-\cos x is much smaller than x. In fact 1-\cos x\approx\dfrac{x^{2}}{2}, so \lim_{x\to0}\dfrac{1-\cos x}{x^{2}}=\dfrac12.

\dfrac{\sin x}{x} is not defined at 0, but its graph heads to the height 1 from both sides.

Results that follow from the sine limit

LimitReason
\lim_{x\to0}\dfrac{\sin ax}{x}=awrite \dfrac{\sin ax}{x}=a\cdot\dfrac{\sin ax}{ax} and let u=ax\to0
\lim_{x\to0}\dfrac{x}{\sin x}=1reciprocal of a limit equal to 1
\lim_{x\to0}\dfrac{\tan x}{x}=1\dfrac{\tan x}{x}=\dfrac{\sin x}{x}\cdot\dfrac{1}{\cos x}\to1\cdot1
\lim_{x\to0}\dfrac{\sin ax}{\sin bx}=\dfrac ab\dfrac{\sin ax}{ax}\cdot\dfrac{bx}{\sin bx}\cdot\dfrac{a}{b}
\lim_{x\to0}\dfrac{1-\cos ax}{x^{2}}=\dfrac{a^{2}}{2}1-\cos ax=2\sin^{2}\dfrac{ax}{2}
Example: using the sine limit
Evaluate (a) \lim_{x\to0}\dfrac{\sin5x}{\sin2x} (b) \lim_{x\to0}\dfrac{1-\cos4x}{x^{2}}
  1. (a) Create \dfrac{\sin5x}{5x} and \dfrac{2x}{\sin2x}, each of which tends to 1.
    \dfrac{\sin5x}{\sin2x}=\dfrac{\sin5x}{5x}\cdot\dfrac{2x}{\sin2x}\cdot\dfrac{5x}{2x}\ \to\ 1\cdot1\cdot\dfrac52=\dfrac52
  2. (b) Use 1-\cos4x=2\sin^{2}2x, then make \dfrac{\sin2x}{2x} appear.
    \dfrac{1-\cos4x}{x^{2}}=\dfrac{2\sin^{2}2x}{x^{2}}=2\cdot4\left(\dfrac{\sin2x}{2x}\right)^{2}\ \to\ 8
\text{(a) }\tfrac52\qquad\text{(b) }8

Limits that define the number e

The number e\approx2.71828 is the limit of \left(1+\dfrac1n\right)^{n} as n grows. These limits are not in the new book's text, but they are standard and FBISE papers use them (2026 Curriculum 2000 paper, Q.2 (vi) OR).

n110100100010000
\left(1+\frac1n\right)^{n}22.59372.70482.71692.7181
\lim_{n\to\infty}\left(1+\dfrac1n\right)^{n}=e,\qquad\lim_{x\to0}(1+x)^{1/x}=e
\lim_{x\to\infty}\left(1+\dfrac ax\right)^{bx}=e^{ab},\qquad\lim_{x\to0}(1+ax)^{1/x}=e^{a}

Proof idea: \left(1+\dfrac ax\right)^{bx}=\left[\left(1+\dfrac ax\right)^{x/a}\right]^{ab} and the bracket tends to e.

\lim_{x\to0}\dfrac{e^{x}-1}{x}=1,\qquad\lim_{x\to0}\dfrac{a^{x}-1}{x}=\ln a\ \ (a>0)
Example: limits of the e type
Evaluate (a) \lim_{x\to\infty}\left(1+\dfrac3x\right)^{x} (b) \lim_{x\to0}(1-2x)^{1/x}
  1. (a) Here a=3, b=1. Write the power so that the inside and the exponent match.
    \left(1+\dfrac3x\right)^{x}=\left[\left(1+\dfrac3x\right)^{x/3}\right]^{3}\to e^{3}
  2. (b) Here a=-2. Put u=-2x, so \dfrac1x=\dfrac{-2}{u} and u\to0.
    (1-2x)^{1/x}=\left[(1+u)^{1/u}\right]^{-2}\to e^{-2}
\text{(a) }e^{3}\qquad\text{(b) }e^{-2}

Limits at infinity

\lim_{x\to\infty}f(x)=L means f(x) gets close to L as x becomes very large. The basic fact is \lim_{x\to\infty}\dfrac{1}{x^{n}}=0 for n>0. For a rational function, divide the top and bottom by the highest power of x in the denominator.

\lim_{x\to\infty}\dfrac{3x^{2}+1}{2x^{2}-5}=\lim_{x\to\infty}\dfrac{3+\frac{1}{x^{2}}}{2-\frac{5}{x^{2}}}=\dfrac{3+0}{2-0}=\dfrac32

2.2, 2.2.1Continuity at a numberOften examined

A function is continuous when its graph has no gaps, holes or jumps: you can draw it without lifting the pencil from the paper. The figure shows the five things that can happen at a number, here x=1.

Only in (v) is the function continuous at 1. Each of (i) to (iv) breaks at least one of the three conditions below.
Continuity at a number
A function f is continuous at a number a if all three conditions hold: (i) f(a) is defined; (ii) \lim_{x\to a}f(x) exists; (iii) \lim_{x\to a}f(x)=f(a). If any one condition fails, f is discontinuous at a.
Figuref(1) definedlimit existslimit =f(1)continuous at 1
(i)nononono
(ii)yesnonono
(iii)noyesnono
(iv)yesyesnono
(v)yesyesyesyes
Example 5 (book)
Where is f(x)=\dfrac{x^{3}-1}{x-1} discontinuous?
  1. The denominator is zero at x=1, so f(1) is not defined. Condition (i) fails, so f is discontinuous at 1.
  2. The limit still exists. Factorise the difference of cubes and cancel x-1 (allowed for x\neq1).
    \lim_{x\to1}\dfrac{(x-1)(x^{2}+x+1)}{x-1}=\lim_{x\to1}(x^{2}+x+1)=3
  3. At every other number f agrees with the polynomial x^{2}+x+1, so f is continuous at every x\neq1.
\text{discontinuous only at } x=1
Example 6 (book)
Is f(x)=\begin{cases}x^{2}, & x<2\\ 5, & x=2\\ -x+6, & x>2\end{cases} continuous at 2?
  1. Condition (i): f(2) is given by the middle rule.
    f(2)=5
  2. Condition (ii): find both one-sided limits from the pieces on each side of 2.
    \lim_{x\to2^-}x^{2}=4,\qquad\lim_{x\to2^+}(-x+6)=4\ \Longrightarrow\ \lim_{x\to2}f(x)=4
  3. Condition (iii) fails because the limit is not equal to the value.
    \lim_{x\to2}f(x)=4\neq5=f(2)
f\ \text{is discontinuous at } 2
Example 7 (book): removing a discontinuity
Let f(x)=\dfrac{x^{2}+x-6}{x^{2}-4} for x\neq2. How should f(2) be defined to make f continuous at 2?
  1. Factorise and cancel x-2 (allowed for x\neq2).
    f(x)=\dfrac{(x-2)(x+3)}{(x-2)(x+2)}=\dfrac{x+3}{x+2},\qquad x\neq2
  2. Find the limit at 2.
    \lim_{x\to2}f(x)=\dfrac{2+3}{2+2}=\dfrac54
  3. Define f(2) equal to this limit. Then all three conditions hold. The new function is called the continuous extension of f.
    f(x)=\begin{cases}\dfrac{x^{2}+x-6}{x^{2}-4}, & x\neq2\\ \dfrac54, & x=2\end{cases}
Before the extension there is a hole at (2,\tfrac54). Defining f(2)=\tfrac54 fills it.
f(2)=\dfrac54

Kinds of discontinuity

Removable discontinuity
\lim_{x\to a}f(x) exists, but f(a) is not defined or f(a)\neq\lim_{x\to a}f(x). Defining (or redefining) f(a) as the limit removes it. Example: \dfrac{x^{2}-1}{x-1} at 1; setting f(1)=2 makes it continuous everywhere.
Jump discontinuity
Both one-sided limits exist but are different, so the graph jumps (Example 1, \dfrac{|x|}{x} at 0). It cannot be removed.
Infinite discontinuity
f(x) grows without bound near a; the graph has a vertical asymptote at x=a, for example \dfrac1{x-3} at 3.

How to test a piecewise function at a joining point

  1. Find f(a) from the piece whose condition contains the equality sign at a (\le, \ge or =).
  2. Find \lim_{x\to a^-}f(x) using the piece that holds just to the left of a.
  3. Find \lim_{x\to a^+}f(x) using the piece that holds just to the right of a.
  4. If all three numbers are equal, f is continuous at a. If unknown constants appear, set the three expressions equal and solve.
Example: two unknown constants (exam type)
Find a and b so that f(x)=\begin{cases}ax+1, & x\le1\\ x^{2}+b, & 1<x<2\\ 3x-a, & x\ge2\end{cases} is continuous at x=1 and x=2.
  1. **At x=1.** The left piece gives f(1) and the left limit; the middle piece gives the right limit. Continuity needs them equal.
    a(1)+1=1^{2}+b\ \Longrightarrow\ a-b=0\qquad(1)
  2. **At x=2.** The middle piece gives the left limit; the last piece gives f(2) and the right limit.
    2^{2}+b=3(2)-a\ \Longrightarrow\ a+b=2\qquad(2)
  3. Solve (1) and (2) together: adding them gives 2a=2.
    a=1,\qquad b=1
  4. Check. With a=b=1: at x=1 both sides give 2; at x=2 both sides give 5.
    f(1)=2=\lim_{x\to1^+}(x^{2}+1),\qquad\lim_{x\to2^-}(x^{2}+1)=5=f(2)
With a=b=1 the three pieces join without a gap.
a=1,\quad b=1

2.2.2Continuity on an interval

Continuous on an open interval
f is continuous on (a,b) if it is continuous at every number in (a,b).
Continuous on a closed interval
f is continuous on [a,b] if it is continuous on (a,b) and, in addition, \lim_{x\to a^+}f(x)=f(a) (continuous from the right at a) and \lim_{x\to b^-}f(x)=f(b) (continuous from the left at b).

At an endpoint we only look from inside the interval. Half-open intervals such as [a,b) or [a,\infty) are treated the same way: check the one-sided limit at each included endpoint.

Example 8 (book)
Discuss the continuity of (a) f(x)=\dfrac{1}{\sqrt{1-x^{2}}} on (-1,1) and on [-1,1] (b) f(x)=\sqrt{1-x^{2}} on [-1,1] (c) f(x)=\sqrt{x-1} on [1,\infty)
  1. (a) For -1<x<1, 1-x^{2}>0, so f is defined and continuous on (-1,1). But f(-1) and f(1) are not defined (division by zero), so f is not continuous on [-1,1].
  2. (b) f is continuous on (-1,1) and at the endpoints the one-sided limits equal the values, so f is continuous on [-1,1].
    \lim_{x\to-1^+}\sqrt{1-x^{2}}=0=f(-1),\qquad\lim_{x\to1^-}\sqrt{1-x^{2}}=0=f(1)
  3. (c) f is continuous for x>1, and at the included endpoint:
    \lim_{x\to1^+}\sqrt{x-1}=0=f(1)\ \Longrightarrow\ \text{continuous on }[1,\infty)
(a) blows up at x=\pm1; (b) is a semicircle that includes its endpoints.
(c) starts at (1,0) and is continuous from the right there.

Which functions are continuous (key facts)

  • If f and g are continuous at a, then cf (c a constant), f+g, f-g, fg and \dfrac fg (when g(a)\neq0) are continuous at a. A composition of continuous functions is continuous.
  • Polynomials, \sin x, \cos x and e^{x} are continuous for every real x.
  • A rational function is continuous everywhere except where its denominator is zero.
  • \tan x and \sec x are discontinuous where \cos x=0, that is, at odd multiples of \dfrac{\pi}{2}; \cot x and \operatorname{cosec}x are discontinuous where \sin x=0, that is, at multiples of \pi.
  • \ln x is continuous for x>0; \sqrt{x} is continuous for x\ge0 (from the right at 0).

Intermediate value theorem

The book does not state this theorem, but Exercise 2.2 Q.19 needs it, and it is the reason the root-finding methods of Unit 10 work ("a root lies where the sign changes").

Intermediate value theorem
If f is continuous on [a,b] and N is any number between f(a) and f(b), then there is at least one c in (a,b) with f(c)=N. In particular, if f(a) and f(b) have opposite signs, the equation f(x)=0 has a root in (a,b).
Example: locating a root
Show that x^{3}+x-1=0 has a root between 0 and 1.
  1. Let f(x)=x^{3}+x-1. It is a polynomial, so it is continuous on [0,1].
  2. Check the signs at the endpoints.
    f(0)=-1<0,\qquad f(1)=1+1-1=1>0
  3. The value 0 lies between f(0) and f(1), so by the intermediate value theorem there is c in (0,1) with f(c)=0. (Numerically c\approx0.682.)
A continuous curve that is below the axis at one end and above it at the other must cross the axis.

2.3Tangent to a graph and rate of change

To write the equation of the tangent line L to the curve y=f(x) at a point P we need two things: the coordinates of P and the slope m_{\tan} of L. The point is easy: if x=a, then P is (a,f(a)). The slope needs a limit.

Take a second point Q on the curve with x-coordinate a+\Delta x, where \Delta x (read “delta x”) is a small change in x. The line through P and Q is a secant. Its slope is the change in y divided by the change in x:

m_{\sec}=\dfrac{\text{change in }y}{\text{change in }x}=\dfrac{f(a+\Delta x)-f(a)}{(a+\Delta x)-a}=\dfrac{\Delta y}{\Delta x}
As \Delta x\to0, Q slides along the curve towards P and the secant PQ turns into the tangent at P.
Tangent line
Let y=f(x) be continuous. The tangent line to the graph at (a,f(a)) is the line through this point with slope m_{\tan}=\lim_{\Delta x\to0}\dfrac{f(a+\Delta x)-f(a)}{\Delta x}=\lim_{\Delta x\to0}\dfrac{\Delta y}{\Delta x}, whenever this limit exists. This slope is also called the slope of the curve at the point.

The four-step method

  1. Evaluate f(a) and f(a+\Delta x).
  2. Find \Delta y=f(a+\Delta x)-f(a) and simplify.
  3. Divide by \Delta x (\Delta x\neq0) and cancel.
  4. Let \Delta x\to0: m_{\tan}=\lim_{\Delta x\to0}\dfrac{\Delta y}{\Delta x}.
Example 9 (book)
Use the definition to find the slope of the tangent to f(x)=x^{2} at (1,f(1)). Also write the equation of the tangent.
  1. Step 1: values at a=1 and 1+\Delta x.
    f(1)=1,\qquad f(1+\Delta x)=(1+\Delta x)^{2}=1+2\Delta x+(\Delta x)^{2}
  2. Step 2: the change in y.
    \Delta y=f(1+\Delta x)-f(1)=2\Delta x+(\Delta x)^{2}=\Delta x(2+\Delta x)
  3. Step 3: divide by \Delta x.
    \dfrac{\Delta y}{\Delta x}=2+\Delta x
  4. Step 4: take the limit.
    m_{\tan}=\lim_{\Delta x\to0}(2+\Delta x)=2
  5. Tangent through (1,1) with slope 2 (point-slope form y-y_1=m(x-x_1)).
    y-1=2(x-1)\ \Longrightarrow\ y=2x-1
m_{\tan}=2,\qquad y=2x-1
Example 10 (book)
Find the slope of the tangent to f(x)=-x^{2}+6x at (4,f(4)).
  1. Step 1.
    \begin{aligned}f(4)&=-16+24=8\\ f(4+\Delta x)&=-(4+\Delta x)^{2}+6(4+\Delta x)\\ &=8-2\Delta x-(\Delta x)^{2}\end{aligned}
  2. Step 2.
    \Delta y=f(4+\Delta x)-f(4)=-2\Delta x-(\Delta x)^{2}=\Delta x(-2-\Delta x)
  3. Step 3.
    \dfrac{\Delta y}{\Delta x}=-2-\Delta x
  4. Step 4.
    m_{\tan}=\lim_{\Delta x\to0}(-2-\Delta x)=-2
  5. The tangent at (4,8) is therefore y-8=-2(x-4), that is y=-2x+16. The negative slope shows the curve is falling at x=4 (it is past its top at x=3).
m_{\tan}=-2

When a tangent does not exist

  • f is discontinuous at x=a (a jump or a hole).
  • The graph has a corner at (a,f(a)), where the slopes from the two sides differ.
  • The graph has a sharp peak (cusp) at the point.
At the corner (2,0) the slope is -1 from the left and +1 from the right, so no single tangent line exists.

Rate of change

The slope \dfrac{\Delta y}{\Delta x} of a secant is the average rate of change of f over the interval from a to a+\Delta x. The slope m_{\tan}=\lim_{\Delta x\to0}\dfrac{\Delta y}{\Delta x} is the instantaneous rate of change of f at a. For example, if m_{\tan}=\dfrac1{10} at (a,f(a)), then near a the value of f changes by only about \dfrac1{10} unit for each unit change in x.

Example: average and instantaneous rates
For f(x)=x^{2} find the average rate of change on [1,3] and the instantaneous rate of change at x=1.
  1. Average rate: change in f divided by change in x.
    \dfrac{f(3)-f(1)}{3-1}=\dfrac{9-1}{2}=4
  2. Instantaneous rate at 1: this is the tangent slope found in Example 9.
    m_{\tan}=2
\text{average }4,\quad\text{instantaneous }2

2.4Instantaneous velocity

If a bus covers 60 miles in one hour, its average velocity is 60 miles per hour. It did not move at exactly this speed all the time; it slowed down in towns and went faster on open roads. The average does not tell us the velocity at a particular instant.

v_{\text{ave}}=\dfrac{\text{distance travelled}}{\text{time of travel}}

A runner finishes a 10\,\mathrm{km} race in 1.25\,\mathrm{h}, so v_{\text{ave}}=\dfrac{10}{1.25}=8\,\mathrm{km\,h^{-1}}. In the first half hour she runs 5\,\mathrm{km}: v_{\text{ave}}=\dfrac{5}{0.5}=10\,\mathrm{km\,h^{-1}}. From 0.5\,\mathrm{h} to 0.6\,\mathrm{h} she goes from the 5\,\mathrm{km} mark to the 5.7\,\mathrm{km} mark: v_{\text{ave}}=\dfrac{5.7-5}{0.6-0.5}=7\,\mathrm{km\,h^{-1}}. Shrinking the time interval gives an ever better picture of her velocity at t=0.5\,\mathrm{h}. The limit of this process is the instantaneous velocity.

Instantaneous velocity
Let s=f(t) give the position at time t of an object moving in a straight line. The instantaneous velocity at time t_1 is v(t_1)=\lim_{\Delta t\to0}\dfrac{f(t_1+\Delta t)-f(t_1)}{\Delta t}=\lim_{\Delta t\to0}\dfrac{\Delta s}{\Delta t}, whenever the limit exists. Its SI unit is \mathrm{m\,s^{-1}}.
Example 11 (book)
The height (in metres) of a ball dropped from a tower is s=-4.9t^{2}+192, where t is in seconds. Find the instantaneous velocity at t_1=3\,\mathrm{s}.
  1. Step 1: position at 3 and at 3+\Delta t.
    \begin{aligned}f(3)&=-4.9(9)+192=147.9\\ f(3+\Delta t)&=-4.9(\Delta t)^{2}-29.4\Delta t+147.9\end{aligned}
  2. Step 2: change in position.
    \Delta s=f(3+\Delta t)-f(3)=\Delta t(-4.9\Delta t-29.4)
  3. Step 3: average velocity over the short interval.
    \dfrac{\Delta s}{\Delta t}=-4.9\Delta t-29.4
  4. Step 4: limit.
    v(3)=\lim_{\Delta t\to0}(-4.9\Delta t-29.4)=-29.4\,\mathrm{m\,s^{-1}}
Height of the falling ball. The slope of the tangent at t=3 is the velocity at that instant.
v(3)=-29.4\,\mathrm{m\,s^{-1}}

2.5The derivative and differentiation from first principlesOften examined

The same limit appeared twice: as the slope of a tangent and as an instantaneous velocity. It is so important that it gets its own name and symbol.

Derivative
The derivative of f with respect to x is the function f' given by f'(x)=\lim_{\Delta x\to0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}. Its domain is every x in the domain of f for which the limit exists.
Differentiable
f is differentiable at x if this limit exists (as a finite number).
Differentiation
The process of finding the derivative. Finding it straight from the limit is called differentiation from first principles (or by definition, or the \Delta-method).
f'(x)=\dfrac{dy}{dx}=\lim_{\Delta x\to0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}=\lim_{\Delta x\to0}\dfrac{\Delta y}{\Delta x}

At a particular number x_0, f'(x_0) is the slope of the tangent at (x_0,f(x_0)) and the instantaneous rate of change of y with respect to x at x_0.

\dfrac{d}{dx} is an operator: its input is a function y=f(x) and its output is the derivative.
NotationRead asUsed by
f'(x), y'“f prime of x”, “y prime”Newton (Lagrange's prime)
\dfrac{dy}{dx}, \dfrac{df}{dx}, \dfrac{d}{dx}f(x)“the derivative of y (or f) with respect to x”Leibniz
D_xf, Dy“D of f”operator notation

Method

  1. Write f(x) and f(x+\Delta x).
  2. Find \Delta y=f(x+\Delta x)-f(x) and simplify so that a factor \Delta x appears.
  3. Divide by \Delta x and cancel.
  4. Let \Delta x\to0.

Derivation Example 12 (book): derivative of x^{2} by definition

Aim. Show that \dfrac{d}{dx}x^{2}=2x.

  1. Write the definition with f(x)=x^{2}.
    f'(x)=\lim_{\Delta x\to0}\dfrac{(x+\Delta x)^{2}-x^{2}}{\Delta x}
  2. Expand the square; the x^{2} terms cancel.
    =\lim_{\Delta x\to0}\dfrac{x^{2}+2x\Delta x+(\Delta x)^{2}-x^{2}}{\Delta x}=\lim_{\Delta x\to0}\dfrac{2x\Delta x+(\Delta x)^{2}}{\Delta x}
  3. Cancel the common factor \Delta x (allowed because \Delta x\neq0), then let \Delta x\to0.
    =\lim_{\Delta x\to0}(2x+\Delta x)=2x
Result\dfrac{d}{dx}x^{2}=2x

The slope of y=x^{2} at any point is twice its x-coordinate: 2 at x=1, -2 at x=-1 and 0 at the bottom of the parabola.

Example 13 (book)
Find the derivative of y=f(x)=-x^{2}+4x+1 by definition.
  1. Find \Delta y. The terms without \Delta x cancel.
    \Delta y=\left[-(x+\Delta x)^{2}+4(x+\Delta x)+1\right]-\left[-x^{2}+4x+1\right]=\Delta x\left[-2x-\Delta x+4\right]
  2. Divide by \Delta x and take the limit.
    f'(x)=\lim_{\Delta x\to0}\dfrac{\Delta x[-2x-\Delta x+4]}{\Delta x}=\lim_{\Delta x\to0}(-2x-\Delta x+4)=-2x+4
f'(x)=-2x+4

Derivation Example 14 (book): derivative of \sqrt{x} by definition

Aim. Show that \dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}} for x>0, and find the slope of the tangent at x=9.

  1. Write the definition with f(x)=\sqrt{x}.
    f'(x)=\lim_{\Delta x\to0}\dfrac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}
  2. Substituting \Delta x=0 gives \dfrac00. Rationalise: multiply top and bottom by the conjugate \sqrt{x+\Delta x}+\sqrt{x}.
    =\lim_{\Delta x\to0}\dfrac{(x+\Delta x)-x}{\Delta x\left(\sqrt{x+\Delta x}+\sqrt{x}\right)}
  3. Cancel \Delta x, then let \Delta x\to0.
    =\lim_{\Delta x\to0}\dfrac{1}{\sqrt{x+\Delta x}+\sqrt{x}}=\dfrac{1}{2\sqrt{x}}
  4. Slope at x=9: substitute into the derivative.
    \left.\dfrac{dy}{dx}\right|_{x=9}=\dfrac{1}{2\sqrt9}=\dfrac16
Result\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}},\qquad m_{\tan}(9)=\dfrac16

The slope of y=\sqrt x is always positive but gets smaller as x grows, so the curve rises more and more slowly.

Derivation Derivative of \dfrac1x by definition

Aim. Show that \dfrac{d}{dx}\left(\dfrac1x\right)=-\dfrac{1}{x^{2}} for x\neq0.

  1. Write \Delta y and combine the fractions over a common denominator.
    \Delta y=\dfrac{1}{x+\Delta x}-\dfrac1x=\dfrac{x-(x+\Delta x)}{x(x+\Delta x)}=\dfrac{-\Delta x}{x(x+\Delta x)}
  2. Divide by \Delta x.
    \dfrac{\Delta y}{\Delta x}=\dfrac{-1}{x(x+\Delta x)}
  3. Let \Delta x\to0.
    \dfrac{dy}{dx}=\lim_{\Delta x\to0}\dfrac{-1}{x(x+\Delta x)}=-\dfrac{1}{x^{2}}
Result\dfrac{d}{dx}\left(\dfrac1x\right)=-\dfrac1{x^{2}}

The slope of the hyperbola y=\dfrac1x is negative everywhere: the curve falls on both of its branches.

2.6, 2.6.1Power, constant and sum rulesOften examined

Differentiating f(x)=5x^{100}+x^{7/5} from first principles would take a long time. The rules below give derivatives quickly. Each is proved once from the definition and then used freely.

Derivation Theorem 2.1: the power rule for a positive integer n

Aim. Prove that \dfrac{d}{dx}x^{n}=nx^{n-1} when n is a positive integer.

Assumptions.
  • The binomial theorem: (x+\Delta x)^{n}=x^{n}+nx^{n-1}\Delta x+\dfrac{n(n-1)}{2!}x^{n-2}(\Delta x)^{2}+\dots+(\Delta x)^{n}.
  1. Let f(x)=x^{n} and expand f(x+\Delta x) by the binomial theorem.
    f(x+\Delta x)=x^{n}+nx^{n-1}\Delta x+\dfrac{n(n-1)}{2!}x^{n-2}(\Delta x)^{2}+\dots+(\Delta x)^{n}
  2. Subtract f(x)=x^{n}; every remaining term contains \Delta x, so take it out as a factor.
    \Delta y=\Delta x\left[nx^{n-1}+\dfrac{n(n-1)}{2!}x^{n-2}\Delta x+\dots+(\Delta x)^{n-1}\right]
  3. Divide by \Delta x.
    \dfrac{\Delta y}{\Delta x}=nx^{n-1}+\dfrac{n(n-1)}{2!}x^{n-2}\Delta x+\dots+(\Delta x)^{n-1}
  4. Let \Delta x\to0. Every term after the first still contains \Delta x, so it tends to 0.
    \dfrac{d}{dx}x^{n}=\lim_{\Delta x\to0}\dfrac{\Delta y}{\Delta x}=nx^{n-1}
Result\dfrac{d}{dx}x^{n}=nx^{n-1}

To differentiate a power, bring the power down as a multiplier and reduce the power by one.

The power rule is true for every real n (negative, fractional or irrational), wherever x^{n} is defined. The book states this without proof.

Example 15 (book): the power rule
Differentiate x^{4}, x^{7}, x^{50}, x^{200}, x^{31}, \sqrt x, x^{4/5}, \dfrac1x, \dfrac{1}{x^{50}} and 10x^{1/3}.
\begin{aligned}\dfrac{d}{dx}x^{4}&=4x^{3}\\ \dfrac{d}{dx}x^{7}&=7x^{6}\\ \dfrac{d}{dx}x^{50}&=50x^{49}\\ \dfrac{d}{dx}x^{200}&=200x^{199}\\ \dfrac{d}{dx}x^{31}&=31x^{30}\end{aligned}
  1. Rewrite roots and reciprocals as powers first, then apply the rule.
    \dfrac{d}{dx}\sqrt{x}=\dfrac{d}{dx}x^{1/2}=\dfrac12x^{-1/2}=\dfrac{1}{2\sqrt x}
  2. \dfrac{d}{dx}x^{4/5}=\dfrac45x^{-1/5}=\dfrac{4}{5x^{1/5}},\qquad\dfrac{d}{dx}\dfrac1x=\dfrac{d}{dx}x^{-1}=-x^{-2}=-\dfrac1{x^{2}}
  3. \dfrac{d}{dx}\dfrac1{x^{50}}=\dfrac{d}{dx}x^{-50}=-50x^{-51}=-\dfrac{50}{x^{51}},\qquad\dfrac{d}{dx}10x^{1/3}=\dfrac{10}{3}x^{-2/3}=\dfrac{10}{3x^{2/3}}

Constant, constant multiple and sum rules

\dfrac{d}{dx}[c]=0,\qquad\dfrac{d}{dx}[cf(x)]=cf'(x),\qquad\dfrac{d}{dx}[f(x)\pm g(x)]=f'(x)\pm g'(x)

In words: the derivative of a constant is zero (its graph is a horizontal line, slope 0); a constant factor stays outside; and the derivative of a sum or difference is the sum or difference of the derivatives.

Derivation Proof of the sum rule

Aim. Prove that \dfrac{d}{dx}[f(x)+g(x)]=f'(x)+g'(x) when f and g are differentiable.

  1. Let y=f(x)+g(x) and find \Delta y; group the f terms and the g terms.
    \Delta y=[f(x+\Delta x)+g(x+\Delta x)]-[f(x)+g(x)]=[f(x+\Delta x)-f(x)]+[g(x+\Delta x)-g(x)]
  2. Divide by \Delta x.
    \dfrac{\Delta y}{\Delta x}=\dfrac{f(x+\Delta x)-f(x)}{\Delta x}+\dfrac{g(x+\Delta x)-g(x)}{\Delta x}
  3. Take limits using the sum theorem for limits; each fraction tends to a derivative.
    \dfrac{dy}{dx}=f'(x)+g'(x)
Result\dfrac{d}{dx}[f(x)+g(x)]=f'(x)+g'(x)

A polynomial can be differentiated term by term. The same proof with minus signs gives the difference rule.

Example 16 (book)
Differentiate (i) 2x^{6}+x^{-9} (ii) 4x^{5}-\dfrac12x^{4}+9x^{3}+7 (iii) 5x^{-1}-\dfrac15x
  1. (i) Sum rule, then the power rule on each term.
    \dfrac{d}{dx}[2x^{6}+x^{-9}]=2(6x^{5})+(-9)x^{-10}=12x^{5}-\dfrac{9}{x^{10}}
  2. (ii) The derivative of the constant 7 is 0.
    \dfrac{d}{dx}\left[4x^{5}-\tfrac12x^{4}+9x^{3}+7\right]=20x^{4}-2x^{3}+27x^{2}
  3. (iii)
    \dfrac{d}{dx}\left[5x^{-1}-\tfrac15x\right]=-5x^{-2}-\dfrac15=-\dfrac{5}{x^{2}}-\dfrac15
Example 17 (book): simplify first
Find \dfrac{dy}{dx} for (a) y=(x+1)^{2} (b) y=(x+1)(x-2) (c) y=\dfrac{x^{3}+x^{2}}{3x}
  1. (a) Expand, then differentiate term by term.
    y=x^{2}+2x+1\ \Longrightarrow\ \dfrac{dy}{dx}=2x+2
  2. (b) Expand.
    y=x^{2}-x-2\ \Longrightarrow\ \dfrac{dy}{dx}=2x-1
  3. (c) Divide each term of the numerator by 3x (for x\neq0).
    y=\dfrac{x^{2}}{3}+\dfrac{x}{3}\ \Longrightarrow\ \dfrac{dy}{dx}=\dfrac{2}{3}x+\dfrac13

Science and business formulas use other letters. The rules are the same; only the names change.

FunctionDerivativeFunctionDerivative
V(t)=4tV'(t)=\dfrac{dV}{dt}=4H(z)=\dfrac14z^{6}H'(z)=\dfrac{dH}{dz}=\dfrac32z^{5}
A(r)=\pi r^{2}A'(r)=\dfrac{dA}{dr}=2\pi rr(\theta)=4\theta^{3}-3\thetar'(\theta)=\dfrac{dr}{d\theta}=12\theta^{2}-3

2.7Product and quotient rulesOften examined

The derivative of a product is not the product of the derivatives. With f(x)=x^{2} and g(x)=x^{3}: f'(x)g'(x)=(2x)(3x^{2})=6x^{3}, but f(x)g(x)=x^{5} has derivative 5x^{4}\neq6x^{3}.

Derivation The product rule

Aim. Prove that \dfrac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x) when f and g are differentiable.

  1. Let y=f(x)g(x) and write the change in y.
    \Delta y=f(x+\Delta x)\,g(x+\Delta x)-f(x)\,g(x)
  2. Subtract and add the same quantity f(x+\Delta x)g(x), so that differences of f and of g appear.
    \Delta y=f(x+\Delta x)\left[g(x+\Delta x)-g(x)\right]+g(x)\left[f(x+\Delta x)-f(x)\right]
  3. Divide by \Delta x.
    \dfrac{\Delta y}{\Delta x}=f(x+\Delta x)\,\dfrac{g(x+\Delta x)-g(x)}{\Delta x}+g(x)\,\dfrac{f(x+\Delta x)-f(x)}{\Delta x}
  4. Let \Delta x\to0. The fractions tend to g'(x) and f'(x); and f(x+\Delta x)\to f(x) because a differentiable function is continuous (section 2.8).
    \dfrac{dy}{dx}=f(x)g'(x)+g(x)f'(x)
Result\dfrac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)

The first function times the derivative of the second, plus the second function times the derivative of the first.

Example 18 (book): two methods
Find \dfrac{dy}{dx} if y=(4x^{2}-1)(7x^{3}+x).
  1. Method I, product rule. First =4x^{2}-1, second =7x^{3}+x.
    \dfrac{dy}{dx}=(4x^{2}-1)\dfrac{d}{dx}(7x^{3}+x)+(7x^{3}+x)\dfrac{d}{dx}(4x^{2}-1)=(4x^{2}-1)(21x^{2}+1)+(7x^{3}+x)(8x)
  2. Expand and collect.
    =84x^{4}+4x^{2}-21x^{2}-1+56x^{4}+8x^{2}=140x^{4}-9x^{2}-1
  3. Method II, multiply first. Expand the product, then differentiate term by term.
    y=28x^{5}-3x^{3}-x\ \Longrightarrow\ \dfrac{dy}{dx}=140x^{4}-9x^{2}-1
\dfrac{dy}{dx}=140x^{4}-9x^{2}-1
Example 19 (book)
Find \dfrac{dy}{dx} if y=(1+x^{3})\sqrt{x}.
  1. Product rule with \dfrac{d}{dx}\sqrt x=\dfrac1{2\sqrt x}.
    \dfrac{dy}{dx}=(1+x^{3})\cdot\dfrac{1}{2\sqrt x}+\sqrt{x}\,(3x^{2})
  2. Write over the common denominator 2\sqrt x, using \sqrt x\cdot3x^{2}=\dfrac{6x^{3}}{2\sqrt x}.
    =\dfrac{1+x^{3}+6x^{3}}{2\sqrt x}=\dfrac{7x^{3}+1}{2\sqrt x}
\dfrac{dy}{dx}=\dfrac{7x^{3}+1}{2\sqrt x}

Derivation The quotient rule

Aim. Prove that \dfrac{d}{dx}\left[\dfrac{f(x)}{g(x)}\right]=\dfrac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^{2}} when g(x)\neq0.

  1. Let y=\dfrac{f(x)}{g(x)} and put \Delta y over one denominator.
    \Delta y=\dfrac{f(x+\Delta x)}{g(x+\Delta x)}-\dfrac{f(x)}{g(x)}=\dfrac{f(x+\Delta x)g(x)-f(x)g(x+\Delta x)}{g(x+\Delta x)\,g(x)}
  2. In the numerator subtract and add f(x)g(x).
    \Delta y=\dfrac{g(x)\left[f(x+\Delta x)-f(x)\right]-f(x)\left[g(x+\Delta x)-g(x)\right]}{g(x+\Delta x)\,g(x)}
  3. Divide by \Delta x.
    \dfrac{\Delta y}{\Delta x}=\dfrac{g(x)\dfrac{f(x+\Delta x)-f(x)}{\Delta x}-f(x)\dfrac{g(x+\Delta x)-g(x)}{\Delta x}}{g(x+\Delta x)\,g(x)}
  4. Let \Delta x\to0; g(x+\Delta x)\to g(x) because g is continuous.
    \dfrac{dy}{dx}=\dfrac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^{2}}
Result\dfrac{d}{dx}\left[\dfrac{f}{g}\right]=\dfrac{gf'-fg'}{g^{2}}

The denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all over the square of the denominator.

Example 20 (book)
Differentiate y=\dfrac{3x^{2}-1}{2x^{3}+5x^{2}+7} with respect to x.
  1. Quotient rule with numerator 3x^{2}-1 (derivative 6x) and denominator 2x^{3}+5x^{2}+7 (derivative 6x^{2}+10x).
    \dfrac{dy}{dx}=\dfrac{(2x^{3}+5x^{2}+7)(6x)-(3x^{2}-1)(6x^{2}+10x)}{(2x^{3}+5x^{2}+7)^{2}}
  2. Expand the numerator: 12x^{4}+30x^{3}+42x-(18x^{4}+30x^{3}-6x^{2}-10x).
    =\dfrac{-6x^{4}+6x^{2}+52x}{(2x^{3}+5x^{2}+7)^{2}}
\dfrac{dy}{dx}=\dfrac{-6x^{4}+6x^{2}+52x}{(2x^{3}+5x^{2}+7)^{2}}
Summary of differentiation rules (book, page 62).
RuleFormula
Constant\dfrac{d}{dx}[c]=0
Constant multiple\dfrac{d}{dx}[cf]=cf'
Sum or difference\dfrac{d}{dx}[f\pm g]=f'\pm g'
Product\dfrac{d}{dx}[fg]=fg'+gf'
Quotient\dfrac{d}{dx}\left[\dfrac fg\right]=\dfrac{gf'-fg'}{g^{2}}
Power\dfrac{d}{dx}x^{n}=nx^{n-1}

2.8Connection between derivative and continuity

  • If a function is differentiable at a point, it is automatically continuous at that point.
  • The reverse is not always true: a function can be continuous at a point and still not be differentiable there, for example at a sharp corner or a cusp.

Derivation Differentiable implies continuous

Aim. If f'(a) exists, show that \lim_{x\to a}f(x)=f(a).

  1. Write the change in f as the difference quotient times \Delta x (true for \Delta x\neq0).
    f(a+\Delta x)-f(a)=\dfrac{f(a+\Delta x)-f(a)}{\Delta x}\cdot\Delta x
  2. Take the limit as \Delta x\to0 using the product theorem. The fraction tends to f'(a) and \Delta x\to0.
    \lim_{\Delta x\to0}\left[f(a+\Delta x)-f(a)\right]=f'(a)\cdot0=0
  3. So f(a+\Delta x)\to f(a), which is the condition for continuity at a (with x=a+\Delta x).
    \lim_{x\to a}f(x)=f(a)
Resultf\ \text{differentiable at }a\ \Longrightarrow\ f\ \text{continuous at }a

A function with a gap or jump at a point cannot have a tangent there, so it cannot be differentiable there.

Example: |x| is continuous but not differentiable at 0
Show that f(x)=|x| is continuous at 0 but f'(0) does not exist.
  1. Continuity: |x|\to0 from both sides and f(0)=0.
    \lim_{x\to0}|x|=0=f(0)
  2. Right-hand derivative: for \Delta x>0, |\Delta x|=\Delta x.
    \lim_{\Delta x\to0^+}\dfrac{|0+\Delta x|-|0|}{\Delta x}=\lim_{\Delta x\to0^+}\dfrac{\Delta x}{\Delta x}=1
  3. Left-hand derivative: for \Delta x<0, |\Delta x|=-\Delta x.
    \lim_{\Delta x\to0^-}\dfrac{|\Delta x|}{\Delta x}=\lim_{\Delta x\to0^-}\dfrac{-\Delta x}{\Delta x}=-1
  4. The one-sided limits differ, so the limit defining f'(0) does not exist.
The graph of |x| has no break at the origin, but it has a corner there.
|x|\ \text{is continuous at }0\ \text{but not differentiable at }0

2.9Derivatives of trigonometric functionsOften examined

We now find the derivatives of the six trigonometric functions. The angle x is always in radians, because the two limits we need are true only in radians:

\lim_{\Delta x\to0}\dfrac{\sin\Delta x}{\Delta x}=1\qquad\text{and}\qquad\lim_{\Delta x\to0}\dfrac{1-\cos\Delta x}{\Delta x}=0

Derivation Derivative of \sin x from first principles

Aim. Show that \dfrac{d}{dx}\sin x=\cos x.

Assumptions.
  • x is in radians.
  • \sin(A+B)=\sin A\cos B+\cos A\sin B.
  1. Start from the definition of the derivative.
    \dfrac{d}{dx}\sin x=\lim_{\Delta x\to0}\dfrac{\sin(x+\Delta x)-\sin x}{\Delta x}
  2. Expand \sin(x+\Delta x) with the addition formula.
    =\lim_{\Delta x\to0}\dfrac{\sin x\cos\Delta x+\cos x\sin\Delta x-\sin x}{\Delta x}
  3. Group the \sin x terms and the \cos x term so that the two standard limits appear.
    =\lim_{\Delta x\to0}\left[\sin x\left(\dfrac{\cos\Delta x-1}{\Delta x}\right)+\cos x\left(\dfrac{\sin\Delta x}{\Delta x}\right)\right]
  4. \sin x and \cos x do not depend on \Delta x, so they come outside the limits. Use the two standard limits.
    =\sin x\,(0)+\cos x\,(1)=\cos x
Result\dfrac{d}{dx}\sin x=\cos x

The slope of the sine curve at any x equals \cos x: it is 1 at x=0, 0 at the peak x=\tfrac{\pi}{2} and -1 at x=\pi.

Derivation Derivative of \cos x from first principles

Aim. Show that \dfrac{d}{dx}\cos x=-\sin x.

Assumptions.
  • \cos(A+B)=\cos A\cos B-\sin A\sin B.
  1. Definition of the derivative.
    \dfrac{d}{dx}\cos x=\lim_{\Delta x\to0}\dfrac{\cos(x+\Delta x)-\cos x}{\Delta x}
  2. Expand with the addition formula.
    =\lim_{\Delta x\to0}\dfrac{\cos x\cos\Delta x-\sin x\sin\Delta x-\cos x}{\Delta x}
  3. Group into the two standard limits.
    =\cos x\lim_{\Delta x\to0}\dfrac{\cos\Delta x-1}{\Delta x}-\sin x\lim_{\Delta x\to0}\dfrac{\sin\Delta x}{\Delta x}
  4. Evaluate.
    =\cos x\,(0)-\sin x\,(1)=-\sin x
Result\dfrac{d}{dx}\cos x=-\sin x

The book says this is shown “in a similar manner”; these are the full steps. The minus sign reflects that \cos x decreases on (0,\pi) where \sin x>0.

Where \sin x has a peak (x=\tfrac{\pi}{2}) its slope is 0, and there \cos x=0. Where \sin x rises most steeply (x=0, 2\pi), \cos x=1.

Derivation Derivatives of \tan x and \sec x

Aim. Show that \dfrac{d}{dx}\tan x=\sec^{2}x and \dfrac{d}{dx}\sec x=\sec x\tan x.

  1. Write \tan x=\dfrac{\sin x}{\cos x} and use the quotient rule.
    \dfrac{d}{dx}\tan x=\dfrac{\cos x\,(\cos x)-\sin x\,(-\sin x)}{\cos^{2}x}
  2. Use \cos^{2}x+\sin^{2}x=1.
    =\dfrac{\cos^{2}x+\sin^{2}x}{\cos^{2}x}=\dfrac{1}{\cos^{2}x}=\sec^{2}x
  3. Write \sec x=\dfrac{1}{\cos x} and use the quotient rule; the derivative of the constant 1 is 0.
    \dfrac{d}{dx}\sec x=\dfrac{\cos x\,(0)-1\,(-\sin x)}{\cos^{2}x}=\dfrac{\sin x}{\cos^{2}x}
  4. Split into two factors.
    =\dfrac{1}{\cos x}\cdot\dfrac{\sin x}{\cos x}=\sec x\tan x
Result\dfrac{d}{dx}\tan x=\sec^{2}x,\qquad\dfrac{d}{dx}\sec x=\sec x\tan x

In the same way, \cot x=\dfrac{\cos x}{\sin x} gives -\operatorname{cosec}^{2}x and \operatorname{cosec} x=\dfrac{1}{\sin x} gives -\dfrac{\cos x}{\sin^{2}x}=-\operatorname{cosec} x\cot x.

Derivatives of the six trigonometric functions.
FunctionDerivativeWith the chain rule (u a function of x)
\sin x\cos x\dfrac{d}{dx}\sin u=\cos u\,\dfrac{du}{dx}
\cos x-\sin x\dfrac{d}{dx}\cos u=-\sin u\,\dfrac{du}{dx}
\tan x\sec^{2}x\dfrac{d}{dx}\tan u=\sec^{2}u\,\dfrac{du}{dx}
\cot x-\operatorname{cosec}^{2}x\dfrac{d}{dx}\cot u=-\operatorname{cosec}^{2}u\,\dfrac{du}{dx}
\sec x\sec x\tan x\dfrac{d}{dx}\sec u=\sec u\tan u\,\dfrac{du}{dx}
\operatorname{cosec} x-\operatorname{cosec} x\cot x\dfrac{d}{dx}\operatorname{cosec} u=-\operatorname{cosec} u\cot u\,\dfrac{du}{dx}
Example 21 (book)
Find \dfrac{dy}{dx} if y=x\sin x.
  1. Product rule.
    \dfrac{dy}{dx}=x\dfrac{d}{dx}\sin x+\sin x\dfrac{d}{dx}x=x\cos x+\sin x
\dfrac{dy}{dx}=x\cos x+\sin x
Example 22 (book)
Find \dfrac{dy}{dx} if y=\dfrac{\sin x}{1+\cos x}.
  1. Quotient rule.
    \dfrac{dy}{dx}=\dfrac{(1+\cos x)\cos x-\sin x(0-\sin x)}{(1+\cos x)^{2}}=\dfrac{\cos x+\cos^{2}x+\sin^{2}x}{(1+\cos x)^{2}}
  2. Use \cos^{2}x+\sin^{2}x=1 and cancel one factor 1+\cos x.
    =\dfrac{1+\cos x}{(1+\cos x)^{2}}=\dfrac{1}{1+\cos x}
\dfrac{dy}{dx}=\dfrac{1}{1+\cos x}
Example 23 (book)
Find \dfrac{dy}{dx} if y=\dfrac{\cos x}{x-\cot x}.
  1. Quotient rule; \dfrac{d}{dx}(x-\cot x)=1+\operatorname{cosec}^{2}x.
    \dfrac{dy}{dx}=\dfrac{(x-\cot x)(-\sin x)-\cos x\,(1+\operatorname{cosec}^{2}x)}{(x-\cot x)^{2}}
  2. Expand. Since \cot x\sin x=\cos x, the terms +\cos x and -\cos x cancel.
    =\dfrac{-x\sin x+\cos x-\cos x-\cos x\operatorname{cosec}^{2}x}{(x-\cot x)^{2}}=\dfrac{-x\sin x-\cos x\operatorname{cosec}^{2}x}{(x-\cot x)^{2}}
\dfrac{dy}{dx}=-\dfrac{x\sin x+\cos x\operatorname{cosec}^{2}x}{(x-\cot x)^{2}}
Example 24 (book)
Find \dfrac{dy}{dx} if y=\sin x(2+\sec x).
  1. Product rule.
    \dfrac{dy}{dx}=\sin x\,(\sec x\tan x)+(2+\sec x)\cos x
  2. Simplify with \sin x\sec x=\tan x and \sec x\cos x=1, then use 1+\tan^{2}x=\sec^{2}x.
    =\tan^{2}x+2\cos x+1=\sec^{2}x+2\cos x
\dfrac{dy}{dx}=\sec^{2}x+2\cos x

2.10Derivatives of inverse trigonometric functionsOften examined

Each inverse trigonometric function is found by the same trick: rewrite y=\sin^{-1}x as x=\sin y, differentiate with respect to x, and express the answer in terms of x. \tan^{-1}x and \cot^{-1}x are differentiable for all real x; the other four are not differentiable at x=\pm1.

Derivation Derivative of \sin^{-1}x

Aim. Show that \dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^{2}}} for -1<x<1.

Assumptions.
  • y=\sin^{-1}x with -\dfrac{\pi}{2}<y<\dfrac{\pi}{2}, where \cos y>0.
  1. Rewrite without the inverse.
    y=\sin^{-1}x\iff x=\sin y
  2. Differentiate both sides with respect to x; the chain rule gives \dfrac{d}{dx}\sin y=\cos y\dfrac{dy}{dx}.
    1=\cos y\,\dfrac{dy}{dx}\ \Longrightarrow\ \dfrac{dy}{dx}=\dfrac{1}{\cos y}
  3. Express \cos y in terms of x. The positive root is taken because \cos y>0 in the range of y.
    \cos y=\sqrt{1-\sin^{2}y}=\sqrt{1-x^{2}}
Result\dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^{2}}},\quad -1<x<1

Similarly \dfrac{d}{dx}\cos^{-1}x=-\dfrac{1}{\sqrt{1-x^{2}}}. The two add to 0, as they should, since \sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2} is a constant.

Derivation Derivative of \tan^{-1}x

Aim. Show that \dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^{2}} for all real x.

Assumptions.
  • y=\tan^{-1}x with -\infty<x<\infty and -\dfrac{\pi}{2}<y<\dfrac{\pi}{2}.
  1. Rewrite without the inverse.
    y=\tan^{-1}x\iff x=\tan y
  2. Differentiate with respect to x.
    1=\sec^{2}y\,\dfrac{dy}{dx}\ \Longrightarrow\ \dfrac{dy}{dx}=\dfrac{1}{\sec^{2}y}
  3. Use \sec^{2}y=1+\tan^{2}y=1+x^{2}.
    \dfrac{dy}{dx}=\dfrac{1}{1+\tan^{2}y}=\dfrac{1}{1+x^{2}}
Result\dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^{2}},\quad x\in\mathbb{R}

Similarly \dfrac{d}{dx}\cot^{-1}x=-\dfrac{1}{1+x^{2}}.

Derivation Derivative of \sec^{-1}x

Aim. Show that \dfrac{d}{dx}\sec^{-1}x=\dfrac{1}{x\sqrt{x^{2}-1}} for |x|>1.

Assumptions.
  • The book takes 0<y<\dfrac{\pi}{2} or \pi<y<\dfrac{3\pi}{2} for y=\sec^{-1}x; in both ranges \tan y>0.
  1. Rewrite without the inverse.
    y=\sec^{-1}x\iff x=\sec y
  2. Differentiate with respect to x.
    1=\sec y\tan y\,\dfrac{dy}{dx}\ \Longrightarrow\ \dfrac{dy}{dx}=\dfrac{1}{\sec y\tan y}
  3. From 1+\tan^{2}y=\sec^{2}y, \tan y=\sqrt{\sec^{2}y-1}=\sqrt{x^{2}-1} (positive root because \tan y>0).
    \dfrac{dy}{dx}=\dfrac{1}{\sec y\sqrt{\sec^{2}y-1}}=\dfrac{1}{x\sqrt{x^{2}-1}}
Result\dfrac{d}{dx}\sec^{-1}x=\dfrac{1}{x\sqrt{x^{2}-1}},\quad |x|>1

Similarly \dfrac{d}{dx}\operatorname{cosec}^{-1}x=-\dfrac{1}{x\sqrt{x^{2}-1}}.

Derivatives of the inverse trigonometric functions (u is a function of x and u'=\dfrac{du}{dx}).
FunctionDerivativeValid forChain rule form
\sin^{-1}x\dfrac{1}{\sqrt{1-x^{2}}}-1<x<1\dfrac{u'}{\sqrt{1-u^{2}}}
\cos^{-1}x-\dfrac{1}{\sqrt{1-x^{2}}}-1<x<1-\dfrac{u'}{\sqrt{1-u^{2}}}
\tan^{-1}x\dfrac{1}{1+x^{2}}all real x\dfrac{u'}{1+u^{2}}
\cot^{-1}x-\dfrac{1}{1+x^{2}}all real x-\dfrac{u'}{1+u^{2}}
\sec^{-1}x\dfrac{1}{x\sqrt{x^{2}-1}}|x|>1\dfrac{u'}{u\sqrt{u^{2}-1}}
\operatorname{cosec}^{-1}x-\dfrac{1}{x\sqrt{x^{2}-1}}|x|>1-\dfrac{u'}{u\sqrt{u^{2}-1}}
Examples 25, 26 and 27 (book)
Differentiate (a) y=\sin^{-1}5x (b) y=\tan^{-1}12x (c) y=\sec^{-1}x^{2} with respect to x.
  1. (a) Chain rule with u=5x, u'=5.
    \dfrac{d}{dx}\sin^{-1}5x=\dfrac{1}{\sqrt{1-(5x)^{2}}}\cdot5=\dfrac{5}{\sqrt{1-25x^{2}}}
  2. (b) u=12x, u'=12.
    \dfrac{d}{dx}\tan^{-1}12x=\dfrac{1}{1+(12x)^{2}}\cdot12=\dfrac{12}{1+144x^{2}}
  3. (c) u=x^{2}, u'=2x; then cancel one x.
    \dfrac{d}{dx}\sec^{-1}x^{2}=\dfrac{1}{x^{2}\sqrt{(x^{2})^{2}-1}}\cdot2x=\dfrac{2x}{x^{2}\sqrt{x^{4}-1}}=\dfrac{2}{x\sqrt{x^{4}-1}}

2.11Chain rule (derivative of a composite function)Often examined

Consider y=(x^{5}+1)^{2}. Writing it as a product (x^{5}+1)(x^{5}+1) and using the product rule gives

\dfrac{dy}{dx}=(x^{5}+1)(5x^{4})+(x^{5}+1)(5x^{4})=2(x^{5}+1)(5x^{4})

The answer has the pattern: bring down the power 2, keep the bracket with power one less, and multiply by the derivative of the inside, 5x^{4}. This is the power rule for functions.

Power rule for functions
If n is a real number and g is differentiable, then \dfrac{d}{dx}[g(x)]^{n}=n[g(x)]^{n-1}g'(x).
Chain rule
If y=f(u) is a differentiable function of u and u=g(x) is a differentiable function of x, then \dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}=f'(g(x))\,g'(x).

Derivation Why the chain rule works

Aim. Show that \dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx} for y=f(u), u=g(x).

Assumptions.
  • For small \Delta x\neq0 the change \Delta u is not zero (a full proof removes this assumption).
  1. A change \Delta x in x causes a change \Delta u=g(x+\Delta x)-g(x) in u, which causes a change \Delta y=f(u+\Delta u)-f(u) in y.
  2. Multiply and divide by \Delta u.
    \dfrac{\Delta y}{\Delta x}=\dfrac{\Delta y}{\Delta u}\cdot\dfrac{\Delta u}{\Delta x}
  3. As \Delta x\to0, \Delta u\to0 because g is continuous, so each fraction tends to a derivative.
    \lim_{\Delta x\to0}\dfrac{\Delta y}{\Delta x}=\lim_{\Delta u\to0}\dfrac{\Delta y}{\Delta u}\cdot\lim_{\Delta x\to0}\dfrac{\Delta u}{\Delta x}
Result\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}

Rates multiply along a chain: if y changes 3 times as fast as u, and u changes 2 times as fast as x, then y changes 6 times as fast as x.

Method: outside, then inside

  1. Identify the outer function and the inner function. In \sin(5x) the outer is \sin(\;) and the inner is 5x.
  2. Differentiate the outer function, keeping the inside unchanged: \cos(5x).
  3. Multiply by the derivative of the inside: \cos(5x)\cdot5.
Example 28 (book)
Differentiate (a) y=(2x^{3}+4x+1)^{4} (b) y=\dfrac{1}{(7x^{5}-x^{4}+2)^{10}}
  1. (a) Power rule for functions with inside 2x^{3}+4x+1.
    \dfrac{dy}{dx}=4(2x^{3}+4x+1)^{3}\cdot(6x^{2}+4)
  2. (b) Write as a negative power first.
    y=(7x^{5}-x^{4}+2)^{-10}\ \Longrightarrow\ \dfrac{dy}{dx}=-10(7x^{5}-x^{4}+2)^{-11}(35x^{4}-4x^{3})=-\dfrac{10(35x^{4}-4x^{3})}{(7x^{5}-x^{4}+2)^{11}}
Example 29 (book): quotient and chain rules together
Differentiate y=\dfrac{(x^{2}-1)^{3}}{(5x+1)^{8}}.
  1. Quotient rule; each bracket is differentiated with the power rule for functions.
    \dfrac{dy}{dx}=\dfrac{(5x+1)^{8}\cdot3(x^{2}-1)^{2}(2x)-(x^{2}-1)^{3}\cdot8(5x+1)^{7}(5)}{(5x+1)^{16}}
  2. Take out the common factor (x^{2}-1)^{2}(5x+1)^{7}.
    =\dfrac{(x^{2}-1)^{2}(5x+1)^{7}\left[6x(5x+1)-40(x^{2}-1)\right]}{(5x+1)^{16}}
  3. Simplify the bracket and cancel (5x+1)^{7}.
    =\dfrac{(x^{2}-1)^{2}(-10x^{2}+6x+40)}{(5x+1)^{9}}
\dfrac{dy}{dx}=\dfrac{(x^{2}-1)^{2}(-10x^{2}+6x+40)}{(5x+1)^{9}}
Example 30 (book)
Differentiate (a) y=\tan^{2}x (b) y=(9x^{3}+1)^{2}\sin5x
  1. (a) \tan^{2}x=(\tan x)^{2}: power rule for functions with inside \tan x.
    \dfrac{dy}{dx}=2\tan x\cdot\sec^{2}x
  2. (b) Product rule; the chain rule gives \dfrac{d}{dx}\sin5x=5\cos5x and \dfrac{d}{dx}(9x^{3}+1)^{2}=2(9x^{3}+1)(27x^{2}).
    \dfrac{dy}{dx}=(9x^{3}+1)^{2}(5\cos5x)+\sin5x\cdot2(9x^{3}+1)(27x^{2})
  3. Take out the common factor 9x^{3}+1.
    =(9x^{3}+1)\left[45x^{3}\cos5x+5\cos5x+54x^{2}\sin5x\right]
Chain-rule results worth memorising.
FunctionDerivative
(ax+b)^{n}na(ax+b)^{n-1}
\sin(ax+b)a\cos(ax+b)
\cos(ax+b)-a\sin(ax+b)
e^{ax+b}ae^{ax+b}
\ln(ax+b)\dfrac{a}{ax+b}
\sqrt{g(x)}\dfrac{g'(x)}{2\sqrt{g(x)}}

2.12Implicit differentiationOften examined

Explicit function
The dependent variable is written alone, in terms of the independent variable only: y=f(x). Example: y=\dfrac13x^{3}+\dfrac43.
Implicit function
x and y are mixed in an equation that is not solved for y. Example: 3y-x^{3}-4=0 defines the same y implicitly.

Often we cannot, or do not want to, solve for y (try 5y^{2}+\sin y=x^{2}). Implicit differentiation finds \dfrac{dy}{dx} directly. Take xy=1. Solving first gives y=\dfrac1x and \dfrac{dy}{dx}=-\dfrac1{x^{2}}. Implicitly:

  1. Differentiate both sides with respect to x. The left side is a product.
    \dfrac{d}{dx}(xy)=\dfrac{d}{dx}(1)\ \Longrightarrow\ x\dfrac{dy}{dx}+y\cdot1=0
  2. Solve for \dfrac{dy}{dx}.
    \dfrac{dy}{dx}=-\dfrac yx
  3. With y=\dfrac1x this agrees with the explicit result.
    \dfrac{dy}{dx}=-\dfrac{1/x}{x}=-\dfrac1{x^{2}}

Method

  1. Differentiate every term on both sides with respect to x.
  2. Whenever you differentiate an expression in y, multiply by \dfrac{dy}{dx} (chain rule): \dfrac{d}{dx}y^{2}=2y\dfrac{dy}{dx}, \dfrac{d}{dx}\sin y=\cos y\dfrac{dy}{dx}. Use the product rule for terms such as xy or x^{3}y^{2}.
  3. Collect all \dfrac{dy}{dx} terms on one side and the rest on the other.
  4. Factor out \dfrac{dy}{dx} and divide.
Example 31 (book)
Use implicit differentiation to find \dfrac{dy}{dx} if 5y^{2}+\sin y=x^{2}.
  1. Differentiate term by term; each y term picks up \dfrac{dy}{dx}.
    10y\dfrac{dy}{dx}+\cos y\dfrac{dy}{dx}=2x
  2. Factor out \dfrac{dy}{dx} and divide.
    (10y+\cos y)\dfrac{dy}{dx}=2x\ \Longrightarrow\ \dfrac{dy}{dx}=\dfrac{2x}{10y+\cos y}
\dfrac{dy}{dx}=\dfrac{2x}{10y+\cos y}
Example: tangent to a circle
Find the slope of the tangent to x^{2}+y^{2}=25 at (3,4) and write the equation of the tangent.
  1. Differentiate implicitly.
    2x+2y\dfrac{dy}{dx}=0\ \Longrightarrow\ \dfrac{dy}{dx}=-\dfrac xy
  2. Substitute the point.
    \left.\dfrac{dy}{dx}\right|_{(3,4)}=-\dfrac34
  3. Point-slope form.
    y-4=-\dfrac34(x-3)\ \Longrightarrow\ 3x+4y=25
The tangent is perpendicular to the radius: the radius has slope 4/3 and the tangent has slope -3/4.
m=-\dfrac34,\qquad 3x+4y=25

2.13, 2.14Derivatives of exponential and logarithmic functionsOften examined

The natural exponential function e^{x} (with e\approx2.71828) has a special property: it is its own derivative. The natural logarithm \ln x=\log_e x is its inverse.

Derivation Derivative of e^{x}

Aim. Show that \dfrac{d}{dx}e^{x}=e^{x}.

Assumptions.
  • The standard limit \lim_{h\to0}\dfrac{e^{h}-1}{h}=1 (section 2.1).
  1. Definition of the derivative with h=\Delta x.
    \dfrac{d}{dx}e^{x}=\lim_{h\to0}\dfrac{e^{x+h}-e^{x}}{h}
  2. Use e^{x+h}=e^{x}e^{h} and take out the factor e^{x}, which does not depend on h.
    =e^{x}\lim_{h\to0}\dfrac{e^{h}-1}{h}
  3. Use the standard limit.
    =e^{x}\cdot1=e^{x}
Result\dfrac{d}{dx}e^{x}=e^{x}

At every point of y=e^{x} the slope equals the height. With the chain rule, \dfrac{d}{dx}e^{3x}=3e^{3x}.

Derivation Derivative of \ln x

Aim. Show that \dfrac{d}{dx}\ln x=\dfrac1x for x>0.

  1. Rewrite without the logarithm.
    y=\ln x\iff x=e^{y}
  2. Differentiate with respect to x (chain rule on e^{y}).
    1=e^{y}\dfrac{dy}{dx}
  3. Solve and use e^{y}=x.
    \dfrac{dy}{dx}=\dfrac{1}{e^{y}}=\dfrac1x
Result\dfrac{d}{dx}\ln x=\dfrac1x

With the chain rule, \dfrac{d}{dx}\ln u=\dfrac{u'}{u}; for example \dfrac{d}{dx}\ln(x^{3}+1)=\dfrac{3x^{2}}{x^{3}+1}.

Derivation Derivatives of a^{x} and \log_a x

Aim. Show that \dfrac{d}{dx}a^{x}=a^{x}\ln a and \dfrac{d}{dx}\log_a x=\dfrac{1}{x\ln a} (a>0, a\neq1).

  1. Write a^{x} as a power of e: since a=e^{\ln a}, a^{x}=e^{x\ln a}.
    \dfrac{d}{dx}a^{x}=\dfrac{d}{dx}e^{x\ln a}=e^{x\ln a}\cdot\ln a
  2. Replace e^{x\ln a} by a^{x}.
    \dfrac{d}{dx}a^{x}=a^{x}\ln a
  3. Change of base: \log_a x=\dfrac{\ln x}{\ln a}, and \ln a is a constant.
    \dfrac{d}{dx}\log_a x=\dfrac{1}{\ln a}\cdot\dfrac1x=\dfrac{1}{x\ln a}
Result\dfrac{d}{dx}a^{x}=a^{x}\ln a,\qquad\dfrac{d}{dx}\log_a x=\dfrac{1}{x\ln a}

For a=e, \ln e=1 and these reduce to the rules for e^{x} and \ln x.

FunctionDerivativeChain rule form
e^{x}e^{x}\dfrac{d}{dx}e^{u}=e^{u}\,u'
a^{x}a^{x}\ln a\dfrac{d}{dx}a^{u}=a^{u}\ln a\,u'
\ln x\dfrac1x\dfrac{d}{dx}\ln u=\dfrac{u'}{u}
\log_a x\dfrac{1}{x\ln a}\dfrac{d}{dx}\log_a u=\dfrac{u'}{u\ln a}
Example 32 (book)
Differentiate y=x^{2}e^{5x} with respect to x.
  1. Product rule; \dfrac{d}{dx}e^{5x}=5e^{5x}.
    \dfrac{dy}{dx}=x^{2}(5e^{5x})+e^{5x}(2x)=5x^{2}e^{5x}+2xe^{5x}=xe^{5x}(5x+2)
\dfrac{dy}{dx}=xe^{5x}(5x+2)
Example 33 (book)
Differentiate y=\ln(4x^{3}+2x^{2}+9) with respect to x.
  1. \dfrac{d}{dx}\ln u=\dfrac{u'}{u} with u=4x^{3}+2x^{2}+9.
    \dfrac{dy}{dx}=\dfrac{12x^{2}+4x}{4x^{3}+2x^{2}+9}=\dfrac{4x(3x+1)}{4x^{3}+2x^{2}+9}
Example 34 (book)
Differentiate y=4^{3x^{2}+5} with respect to x.
  1. Take \ln of both sides and bring the power down.
    \ln y=(3x^{2}+5)\ln4
  2. Differentiate implicitly; \ln4 is a constant.
    \dfrac1y\dfrac{dy}{dx}=\ln4\cdot6x
  3. Multiply by y.
    \dfrac{dy}{dx}=6x\ln4\cdot4^{3x^{2}+5}

Logarithmic differentiation

When the variable is in both the base and the power, as in x^{x}, neither the power rule nor the exponential rule applies. Take logarithms first, then differentiate implicitly. The same trick simplifies long products and quotients.

Example: y=x^{x}
Find \dfrac{dy}{dx} for y=x^{x} (x>0) and its value at x=1.
  1. Take \ln of both sides.
    \ln y=x\ln x
  2. Differentiate; the right side needs the product rule.
    \dfrac1y\dfrac{dy}{dx}=x\cdot\dfrac1x+\ln x=1+\ln x
  3. Multiply by y=x^{x} and substitute x=1.
    \dfrac{dy}{dx}=x^{x}(1+\ln x),\qquad\left.\dfrac{dy}{dx}\right|_{x=1}=1\cdot(1+0)=1
\dfrac{dy}{dx}=x^{x}(1+\ln x)
e^{x} and \ln x are reflections of each other in y=x. At (0,1) the slope of e^{x} is e^{0}=1; at (1,0) the slope of \ln x is \dfrac11=1.

2.14Parametric differentiationOften examined

Parametric equations
A curve given by x=f(t) and y=g(t), where both coordinates depend on a third variable t called the parameter (often time or an angle).

Derivation Derivative of a parametric function

Aim. Find \dfrac{dy}{dx} when x=f(t) and y=g(t).

Assumptions.
  • \dfrac{dx}{dt}\neq0 at the point.
  1. By the chain rule, y depends on x through t.
    \dfrac{dy}{dx}=\dfrac{dy}{dt}\cdot\dfrac{dt}{dx}
  2. \dfrac{dt}{dx} is the reciprocal of \dfrac{dx}{dt}.
    \dfrac{dt}{dx}=\dfrac{1}{dx/dt}
  3. Combine.
    \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}
Result\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt},\qquad\dfrac{dx}{dt}\neq0

Differentiate x and y separately with respect to t and divide. There is no need to eliminate t.

Example 35 (book)
Find \dfrac{dy}{dx} if x=\tan t, y=4t^{3}+1.
  1. Differentiate each with respect to t.
    \dfrac{dy}{dt}=12t^{2},\qquad\dfrac{dx}{dt}=\sec^{2}t
  2. Divide.
    \dfrac{dy}{dx}=\dfrac{12t^{2}}{\sec^{2}t}=12t^{2}\cos^{2}t
\dfrac{dy}{dx}=12t^{2}\cos^{2}t
Example 36 (book)
Find \dfrac{dy}{dx} if x=\dfrac{1-t^{2}}{1+t^{2}}, y=\dfrac{2t}{1+t^{2}}.
  1. Quotient rule for x.
    \dfrac{dx}{dt}=\dfrac{(1+t^{2})(-2t)-(1-t^{2})(2t)}{(1+t^{2})^{2}}=\dfrac{-4t}{(1+t^{2})^{2}}
  2. Quotient rule for y.
    \dfrac{dy}{dt}=\dfrac{(1+t^{2})(2)-2t(2t)}{(1+t^{2})^{2}}=\dfrac{2(1-t^{2})}{(1+t^{2})^{2}}
  3. Divide; the denominators (1+t^{2})^{2} cancel.
    \dfrac{dy}{dx}=\dfrac{2(1-t^{2})}{-4t}=\dfrac{t^{2}-1}{2t}
Since x^{2}+y^{2}=1, this curve is the unit circle. At t=\tfrac12 the point is (\tfrac35,\tfrac45) and \dfrac{dy}{dx}=\dfrac{1/4-1}{1}=-\dfrac34, the same as -\dfrac xy from implicit differentiation.
\dfrac{dy}{dx}=\dfrac{t^{2}-1}{2t}

2.15DifferentialsOften examined

For a small change \Delta x, the secant slope \dfrac{\Delta y}{\Delta x} is close to the tangent slope f'(x). Multiplying by \Delta x gives a simple estimate of the change in y:

\dfrac{\Delta y}{\Delta x}\approx f'(x)\quad\Longrightarrow\quad\Delta y\approx f'(x)\,\Delta x
Differential of x
The increment \Delta x of the independent variable is called the differential of x and is written dx, so dx=\Delta x.
Differential of y
For y=f(x), the quantity f'(x)\,dx is called the differential of y and is written dy: dy=f'(x)\,dx.
\Delta y is the rise of the curve; dy=f'(x)\,dx is the rise of the tangent. For small \Delta x they are nearly equal.

Geometrically, dy is the rise of the tangent line and \Delta y is the rise of the curve over the same run dx. So \Delta y\approx dy, and the approximation improves as dx gets smaller. Dividing dy=f'(x)\,dx by dx shows that the derivative really is a ratio of two differentials: f'(x)=\dfrac{dy}{dx}.

Example 37 (book)
(a) Find \Delta y and dy for y=5x^{2}+4x+1. (b) Compare them for x=6, \Delta x=dx=0.02.
  1. (a) The exact change.
    \Delta y=[5(x+\Delta x)^{2}+4(x+\Delta x)+1]-[5x^{2}+4x+1]=(10x+4)\Delta x+5(\Delta x)^{2}
  2. The differential uses only the derivative f'(x)=10x+4.
    dy=(10x+4)\,dx
  3. So \Delta y and dy differ by exactly 5(\Delta x)^{2}, which is tiny when \Delta x is small.
  4. (b) Substitute x=6, \Delta x=0.02.
    \Delta y=64(0.02)+5(0.02)^{2}=1.282,\qquad dy=64(0.02)=1.28
  5. The difference is 5(0.02)^{2}=0.002.
    \Delta y\approx dy
For y=5x^{2}+4x+1 at x=6: the smaller dx, the better dy approximates \Delta y.
\Delta x=dx\Delta ydy\Delta y-dy
0.16.456.40.05
0.021.2821.280.002
0.0010.0640050.0640.000005

2.16, 2.22Approximations and errors using differentialsOften examined

If we know f and f' at a convenient number x, differentials predict the value at a nearby number x+\Delta x. Since f(x+\Delta x)=f(x)+\Delta y and \Delta y\approx dy:

f(x+\Delta x)\approx f(x)+dy=f(x)+f'(x)\,dx

This is the tangent-line (linear) approximation. It is the sense in which a derivative lets us approximate a function.

Method

  1. Choose the function f and a convenient x near the given number where f(x) is known exactly (a perfect square, 45^{\circ}, and so on).
  2. Find dx=\Delta x = given number minus x. For angles, convert dx to radians.
  3. Compute f(x) and f'(x).
  4. Use f(x+\Delta x)\approx f(x)+f'(x)\,dx.
Example 38 (book)
Find the approximate value of \sqrt{25.4}.
  1. Take f(x)=\sqrt x, x=25 (a perfect square) and dx=0.4.
    dy=\dfrac{1}{2\sqrt x}\,dx
  2. Apply the approximation.
    \sqrt{25.4}\approx\sqrt{25}+\dfrac{1}{2\sqrt{25}}(0.4)=5+\dfrac{0.4}{10}=5.04
\sqrt{25.4}\approx5.04\quad(\text{calculator: }5.0398)
Example: a cube root and a trigonometric value
Approximate (a) \sqrt[3]{27.3} (b) \cos61^{\circ}.
  1. (a) f(x)=x^{1/3}, x=27, dx=0.3, and f'(x)=\dfrac{1}{3x^{2/3}} so f'(27)=\dfrac{1}{27}.
    \sqrt[3]{27.3}\approx3+\dfrac{0.3}{27}=3.0111
  2. (b) f(x)=\cos x, x=60^{\circ}, dx=1^{\circ}=\dfrac{\pi}{180}=0.017453 rad, and f'(x)=-\sin x.
    \cos61^{\circ}\approx\cos60^{\circ}-\sin60^{\circ}\cdot\dfrac{\pi}{180}=0.5-0.8660(0.017453)=0.4849
\text{(a) }3.0111\qquad\text{(b) }0.4849

Absolute, relative and percentage error

When a length is measured with a small possible error dx, any quantity y=f(x) calculated from it inherits an error, approximately dy.

Absolute error
The approximate error in y: dy=f'(x)\,dx (same unit as y).
Relative error
\dfrac{dy}{y}, the error as a fraction of the value (no unit).
Percentage error
\dfrac{dy}{y}\times100\%.
y=kx^{n}\ \Longrightarrow\ \dfrac{dy}{y}=\dfrac{nkx^{n-1}\,dx}{kx^{n}}=n\,\dfrac{dx}{x}

Shortcut: the relative error of a power is the power times the relative error of the measurement. Cube: \dfrac{dV}{V}=3\dfrac{dx}{x}; sphere: \dfrac{dV}{V}=3\dfrac{dr}{r}; square: \dfrac{dA}{A}=2\dfrac{dx}{x}.

Example 49 (book)
A side of a cube is measured to be 30\,\mathrm{cm} with a possible error of \pm0.02\,\mathrm{cm}. What is the approximate maximum possible error in the volume?
Given
  • x = 30\,\mathrm{cm} side
  • dx = \pm0.02\,\mathrm{cm} possible error
To find
  • dV
  1. The exact error would be \Delta V=(x+\Delta x)^{3}-x^{3}. Use the differential as an approximation to it.
    V=x^{3}\ \Longrightarrow\ dV=3x^{2}\,dx
  2. Substitute.
    dV=3(30)^{2}(\pm0.02)=\pm54\,\mathrm{cm^{3}}
  3. Also the relative and percentage errors (not asked in the book, but often asked in exams).
    \dfrac{dV}{V}=3\,\dfrac{dx}{x}=3\left(\dfrac{0.02}{30}\right)=0.002=0.2\%
dV\approx\pm54\,\mathrm{cm^{3}}
Example: area of a circle
The radius of a circle is measured as 5\,\mathrm{cm} with a possible error of 0.1\,\mathrm{cm}. Estimate the error and the percentage error in the area.
  1. A=\pi r^{2}, so dA=2\pi r\,dr.
    dA=2\pi(5)(0.1)=\pi\approx3.14\,\mathrm{cm^{2}}
  2. Percentage error: power 2 times the relative error in r.
    \dfrac{dA}{A}\times100\%=2\left(\dfrac{0.1}{5}\right)\times100\%=4\%
dA\approx3.14\,\mathrm{cm^{2}},\quad4\%

2.17, 2.18Higher order derivatives

The derivative f'(x) is itself a function, so we can differentiate it again. The result is the second derivative.

f''(x)=y''=\dfrac{d^{2}y}{dx^{2}}=\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)=D^{2}y
Example 39 (book)
Find the second derivative of y=x^{3}-2x^{2}.
  1. First derivative.
    \dfrac{dy}{dx}=3x^{2}-4x
  2. Differentiate again.
    \dfrac{d^{2}y}{dx^{2}}=\dfrac{d}{dx}(3x^{2}-4x)=6x-4
y''=6x-4
Example 40 (book)
Find the second derivative of (a) \sin3x (b) (x^{3}+1)^{4} (c) e^{2x}.
  1. (a) Chain rule twice.
    y'=3\cos3x,\qquad y''=-9\sin3x
  2. (b) First derivative by the power rule for functions.
    y'=4(x^{3}+1)^{3}(3x^{2})=12x^{2}(x^{3}+1)^{3}
  3. Second derivative by the product rule, then factorise.
    y''=12\left[x^{2}\cdot3(x^{3}+1)^{2}(3x^{2})+(x^{3}+1)^{3}(2x)\right]=12x(x^{3}+1)^{2}\left[9x^{3}+2(x^{3}+1)\right]=12x(x^{3}+1)^{2}(11x^{3}+2)
  4. (c)
    y'=2e^{2x},\qquad y''=4e^{2x}

Differentiating again gives the third, fourth and in general the nth derivative:

\dfrac{d^{3}y}{dx^{3}}=\dfrac{d}{dx}\left(\dfrac{d^{2}y}{dx^{2}}\right),\qquad\dfrac{d^{4}y}{dx^{4}}=\dfrac{d}{dx}\left(\dfrac{d^{3}y}{dx^{3}}\right),\qquad\dfrac{d^{n}y}{dx^{n}}=\dfrac{d}{dx}\left(\dfrac{d^{n-1}y}{dx^{n-1}}\right)
OrderNotations
1st to 3rdf'(x),\ f''(x),\ f'''(x); \ y',\ y'',\ y'''; \ D_xy,\ D_x^{2}y,\ D_x^{3}y
4th to nthf^{(4)}(x),\dots,f^{(n)}(x); \ y^{(4)},\dots,y^{(n)}; \ D_x^{4}y,\dots,D_x^{n}y
Examples 41 and 42 (book)
(a) Find the first five derivatives of f(x)=2x^{4}-6x^{3}+7x^{2}+5x-10. (b) Find the third derivative of y=\dfrac{1}{x^{3}}.
  1. (a) Differentiate repeatedly.
    \begin{aligned}f'&=8x^{3}-18x^{2}+14x+5\\ f''&=24x^{2}-36x+14\\ f'''&=48x-36\\ f^{(4)}&=48\\ f^{(5)}&=0\end{aligned}
  2. (b) Write y=x^{-3} and use the power rule three times.
    y'=-3x^{-4},\quad y''=12x^{-5},\quad y'''=-60x^{-6}=-\dfrac{60}{x^{6}}
Derivatives of \sin x repeat every four steps; each derivative of e^{ax} multiplies by a.
Functiony'y''y'''y^{(4)}
\sin x\cos x-\sin x-\cos x\sin x
e^{ax}ae^{ax}a^{2}e^{ax}a^{3}e^{ax}a^{4}e^{ax}

Increasing and decreasing functionsOften examined

Increasing function
f is increasing on an interval I if for any x_1<x_2 in I, f(x_1)<f(x_2): the graph rises from left to right.
Decreasing function
f is decreasing on I if for any x_1<x_2 in I, f(x_1)>f(x_2): the graph falls from left to right.

The derivative test

  • If f'(x)>0 for every x in (a,b), then f is increasing on (a,b).
  • If f'(x)<0 for every x in (a,b), then f is decreasing on (a,b).
  • If f'(c)=0, the tangent at c is horizontal: c is a stationary point, where f may change from increasing to decreasing or the reverse.

Method for the intervals

  1. Find f'(x) and factorise it.
  2. Solve f'(x)=0 (and note where f' does not exist). These numbers split the real line into intervals.
  3. Pick a test value in each interval and find the sign of f' there (or use the signs of the factors).
  4. Write the answer: increasing where f'>0, decreasing where f'<0.
Example: intervals of increase and decrease
Find the intervals on which f(x)=2x^{3}-3x^{2}-12x+1 is increasing or decreasing.
  1. Differentiate and factorise.
    f'(x)=6x^{2}-6x-12=6(x^{2}-x-2)=6(x+1)(x-2)
  2. Solve f'(x)=0.
    x=-1\quad\text{or}\quad x=2
  3. Test the sign of f' in each interval.
IntervalTest valuex+1x-2f'(x)f
(-\infty,-1)x=-2--+increasing
(-1,2)x=0+--decreasing
(2,\infty)x=3+++increasing
The curve rises where the derivative is positive and falls where it is negative. The turning points are where the derivative is zero.
\text{increasing on }(-\infty,-1)\text{ and }(2,\infty);\ \text{decreasing on }(-1,2)
Example: at a single point
Is f(x)=x^{3}-3x increasing or decreasing at x=2 and at x=0?
  1. Find the derivative.
    f'(x)=3x^{2}-3
  2. At x=2 the slope is positive.
    f'(2)=12-3=9>0\ \Longrightarrow\ \text{increasing at }x=2
  3. At x=0 the slope is negative.
    f'(0)=-3<0\ \Longrightarrow\ \text{decreasing at }x=0
  4. If the derivative is 0 at the point, the function is neither increasing nor decreasing there; it is a stationary point (use f'' to see whether it is a maximum or minimum).

2.19Absolute extremaOften examined

Suppose f is defined on an interval I. The largest and smallest values of f on I (if they exist) are called the extrema of f. There are two kinds: absolute (global) and relative (local).

Absolute maximum
f(c) is the absolute maximum of f if f(x)\le f(c) for every x in the domain of f.
Absolute minimum
f(c) is the absolute minimum of f if f(x)\ge f(c) for every x in the domain of f.

Absolute extrema are also called global extrema. A function may have both, one, or neither.

Examples 43 and 44 (book)
Find the absolute extrema, if any, of (i) \sin x (ii) x^{2} (iii) \dfrac1x (iv) x^{2} on [1,2] (v) x^{2} on (1,2) (vi) x^{2} on [-1,2].
Function and domainAbsolute maximumAbsolute minimum
\sin x on \mathbb{R}f\!\left(\tfrac{\pi}{2}\right)=1f\!\left(\tfrac{3\pi}{2}\right)=-1
x^{2} on \mathbb{R}none (grows without bound)f(0)=0
\dfrac1x, x\neq0nonenone
x^{2} on [1,2]f(2)=4f(1)=1
x^{2} on (1,2)none (the endpoints are excluded)none
x^{2} on [-1,2]f(2)=4f(0)=0
On [-1,2] the maximum is at an endpoint and the minimum is inside the interval.
Extreme value theorem
A function that is continuous on a closed interval [a,b] always has an absolute maximum and an absolute minimum on that interval.

The theorem needs both conditions. On the open interval (1,2), x^{2} gets as close as we like to 1 and 4 but never reaches them. And \dfrac1x on [-1,1] has no extrema because it is not continuous at 0.

Closed interval method for absolute extrema

  1. Check that f is continuous on [a,b].
  2. Find the critical values of f that lie inside (a,b) (section 2.19.1).
  3. Evaluate f at each critical value and at the endpoints a and b.
  4. The largest of these values is the absolute maximum; the smallest is the absolute minimum.
Example: closed interval method
Find the absolute extrema of f(x)=x^{3}-12x on [-3,5].
  1. Critical values: solve f'(x)=0.
    f'(x)=3x^{2}-12=3(x-2)(x+2)=0\ \Longrightarrow\ x=-2,\ 2\ (\text{both in }[-3,5])
  2. Evaluate at the critical values and the endpoints.
    f(-3)=9,\quad f(-2)=16,\quad f(2)=-16,\quad f(5)=65
  3. Compare.
    \text{abs. max}=65\ \text{at }x=5,\qquad\text{abs. min}=-16\ \text{at }x=2
The local maximum at x=-2 is not the absolute maximum: the endpoint x=5 gives a larger value.
\max=65\ (x=5),\qquad\min=-16\ (x=2)

2.19.1Relative extrema and critical valuesOften examined

A function with no absolute extrema can still have hills and valleys. Near the top of a hill, f(c) is the largest value in a small neighbourhood of c, even if the function goes higher far away.

Relative maximum
f(c) is a relative (local) maximum of f if f(x)\le f(c) for every x in some open interval that contains c.
Relative minimum
f(c) is a relative (local) minimum of f if f(x)\ge f(c) for every x in some open interval that contains c.

Result. If f has a relative extremum at c, then either f'(c)=0 or f'(c) does not exist. At a smooth hill or valley the tangent is horizontal; at a sharp corner there is no tangent.

Critical value
A number c in the domain of f for which f'(c)=0 or f'(c) does not exist. The point (c,f(c)) is a critical point. When f'(c)=0 it is also called a stationary point.
Example 45 (book)
Find the critical values of (a) f(x)=x^{3}-15x+6 (b) f(x)=(x+4)^{2/3} (c) f(x)=\dfrac{x^{2}}{x-1}
  1. (a) f' exists everywhere, so solve f'(x)=0.
    f'(x)=3x^{2}-15=3(x+\sqrt5)(x-\sqrt5)=0\ \Longrightarrow\ x=-\sqrt5,\ \sqrt5
  2. (b) Differentiate with the power rule.
    f'(x)=\dfrac23(x+4)^{-1/3}=\dfrac{2}{3(x+4)^{1/3}}
  3. f'(x) is never 0, but it does not exist at x=-4, and -4 is in the domain of f (f(-4)=0). So x=-4 is a critical value (a cusp).
  4. (c) Quotient rule.
    f'(x)=\dfrac{(x-1)(2x)-x^{2}(1)}{(x-1)^{2}}=\dfrac{x(x-2)}{(x-1)^{2}}
  5. f'(x)=0 at x=0 and x=2. f' does not exist at x=1, but 1 is not in the domain of f, so it is not a critical value.
    \text{critical values: }0,\ 2
(b) The graph has a sharp point at x=-4, where the derivative does not exist; it is a relative minimum.

First derivative test

This test is not in the book, but it is the natural way to classify a critical value, and it works even when f'(c) does not exist.

  • If f' changes from positive to negative as x increases through c, then f(c) is a relative maximum.
  • If f' changes from negative to positive, then f(c) is a relative minimum.
  • If f' does not change sign, f(c) is not an extremum.
Example: first derivative test
Find and classify the relative extrema of f(x)=x^{3}-3x^{2}.
  1. Critical values.
    f'(x)=3x^{2}-6x=3x(x-2)=0\ \Longrightarrow\ x=0,\ 2
  2. Signs: f'>0 on (-\infty,0), f'<0 on (0,2), f'>0 on (2,\infty).
  3. At x=0 the sign changes from + to -: relative maximum. At x=2 it changes from - to +: relative minimum.
    f(0)=0\ (\text{rel. max}),\qquad f(2)=8-12=-4\ (\text{rel. min})
\text{rel. max }0\text{ at }x=0,\quad\text{rel. min }-4\text{ at }x=2

2.20, 2.21Concavity, points of inflection and the second derivative testOften examined

A curve that bends upward like a cup is concave upward; it “holds water”. A curve that bends downward like a cap is concave downward; it “spills water”. On a concave upward arc the slope f' increases, so f''>0; on a concave downward arc the slope decreases, so f''<0.

Test for concavity
Let f'' exist on (a,b). If f''(x)>0 for all x in (a,b), the graph of f is concave upward on (a,b). If f''(x)<0 for all x in (a,b), it is concave downward on (a,b).
Point of inflection
Let f be continuous at c. The point (c,f(c)) is a point of inflection if there is an open interval (a,b) containing c such that the graph is concave upward on (a,c) and concave downward on (c,b), or the reverse. In short, the concavity changes at c.
Example 46 (book)
Find the intervals on which the graph of f(x)=-x^{3}+\dfrac92x^{2} is concave upward and concave downward.
  1. Find f''.
    f'(x)=-3x^{2}+9x,\qquad f''(x)=-6x+9=6\left(-x+\dfrac32\right)
  2. f''(x)>0 when -x+\dfrac32>0, that is x<\dfrac32.
    \text{concave upward on }\left(-\infty,\tfrac32\right)
  3. f''(x)<0 when x>\dfrac32.
    \text{concave downward on }\left(\tfrac32,\infty\right)
  4. The concavity changes at x=\dfrac32, where f\!\left(\tfrac32\right)=-\tfrac{27}{8}+\tfrac{81}{8}=\tfrac{27}{4}. So \left(\tfrac32,\tfrac{27}{4}\right) is a point of inflection.
Example 47 (book)
Find the points of inflection of f(x)=-x^{3}+x^{2}.
  1. Find f'' and solve f''(x)=0 for the possible point.
    f'(x)=-3x^{2}+2x,\qquad f''(x)=-6x+2=0\ \Longrightarrow\ x=\dfrac13
  2. Check that the sign of f'' really changes.
    f''(x)=6\left(-x+\tfrac13\right)>0\ \text{for }x<\tfrac13,\qquad<0\ \text{for }x>\tfrac13
  3. The concavity changes, so there is an inflection point. Its y-coordinate:
    f\!\left(\tfrac13\right)=-\tfrac{1}{27}+\tfrac19=\tfrac{2}{27}
\left(\tfrac13,\ \tfrac{2}{27}\right)
Second derivative test
Let f'(c)=0 and let f'' exist on an interval containing c. (i) If f''(c)>0, then f(c) is a relative minimum. (ii) If f''(c)<0, then f(c) is a relative maximum. (iii) If f''(c)=0, the test fails; use the first derivative test.
Example 48 (book)
Find the critical points of f(x)=x^{4}-x^{2} and the relative extrema by the second derivative test.
  1. Critical values from f'(x)=0.
    f'(x)=4x^{3}-2x=2x(2x^{2}-1)=0\ \Longrightarrow\ x=0,\ \pm\dfrac{\sqrt2}{2}
  2. Second derivative.
    f''(x)=12x^{2}-2
xf''(x)f(x)Conclusion
0-2<00relative maximum
\dfrac{\sqrt2}{2}4>0-\dfrac14relative minimum
-\dfrac{\sqrt2}{2}4>0-\dfrac14relative minimum

Exam template: critical points, classification and absolute extrema

  1. Write f'(x), factorise it, and solve f'(x)=0 inside the given interval (for trigonometric functions list every solution in the interval).
  2. Write f''(x) and find its sign at each critical value; state maximum or minimum.
  3. Find f at each critical value and at the endpoints.
  4. State the absolute maximum and minimum with the x values where they occur.
Example: a trigonometric function on a closed interval
For f(x)=x-2\sin x on [0,2\pi], find the critical points, classify them, and find the absolute extrema.
  1. Critical values.
    f'(x)=1-2\cos x=0\ \Longrightarrow\ \cos x=\tfrac12\ \Longrightarrow\ x=\dfrac{\pi}{3},\ \dfrac{5\pi}{3}
  2. Second derivative test.
    f''(x)=2\sin x:\quad f''\!\left(\tfrac{\pi}{3}\right)=\sqrt3>0\ (\text{min}),\qquad f''\!\left(\tfrac{5\pi}{3}\right)=-\sqrt3<0\ (\text{max})
  3. Values at the critical points and endpoints.
    \begin{aligned}f(0)&=0\\ f\!\left(\tfrac{\pi}{3}\right)&=\tfrac{\pi}{3}-\sqrt3\approx-0.685\\ f\!\left(\tfrac{5\pi}{3}\right)&=\tfrac{5\pi}{3}+\sqrt3\approx6.968\\ f(2\pi)&=2\pi\approx6.283\end{aligned}
  4. Compare all four values.
    \text{abs. max}=\tfrac{5\pi}{3}+\sqrt3\ \text{at }x=\tfrac{5\pi}{3},\qquad\text{abs. min}=\tfrac{\pi}{3}-\sqrt3\ \text{at }x=\tfrac{\pi}{3}

2.22Applications of derivatives

Many real quantities change: the speed of a rocket, inflation of prices, the number of bacteria in a culture, the voltage of a signal. Derivatives measure how fast they change, and they find the largest or smallest value of a quantity (optimisation). Error estimates with differentials (Example 49 of the book) are covered in the section on approximations and errors.

Related rates

When two quantities are linked by an equation and both change with time t, differentiating the equation with respect to t links their rates.

  1. Draw a figure and name the variables.
  2. Write the equation that links them.
  3. Differentiate both sides with respect to t (chain rule).
  4. Substitute the known values and rates only after differentiating.
  5. Answer with units.
Example 50 (book)
A square is expanding with time. How is the rate at which its area increases related to the rate at which a side increases?
  1. Area in terms of the side x.
    A=x^{2}
  2. Differentiate with respect to t; by the chain rule \dfrac{d}{dt}x^{2}=2x\dfrac{dx}{dt}.
    \dfrac{dA}{dt}=2x\,\dfrac{dx}{dt}
\dfrac{dA}{dt}=2x\,\dfrac{dx}{dt}
Example 51 (book)
Air is pumped into a spherical balloon at 20\,\mathrm{ft^{3}/min}. At what rate is the radius changing when the radius is 3\,\mathrm{ft}?
Given
  • \dfrac{dV}{dt} = 20\,\mathrm{ft^{3}/min}
  • r = 3\,\mathrm{ft}
To find
  • \dfrac{dr}{dt} at r=3
  1. Volume of a sphere.
    V=\dfrac43\pi r^{3}
  2. Differentiate with respect to t.
    \dfrac{dV}{dt}=4\pi r^{2}\,\dfrac{dr}{dt}
  3. Substitute \dfrac{dV}{dt}=20 and solve.
    \dfrac{dr}{dt}=\dfrac{20}{4\pi r^{2}}=\dfrac{5}{\pi r^{2}}
  4. At r=3.
    \dfrac{dr}{dt}=\dfrac{5}{9\pi}\approx0.18\,\mathrm{ft/min}
\dfrac{dr}{dt}=\dfrac{5}{9\pi}\approx0.18\,\mathrm{ft/min}

Optimisation

  1. Name the variables and write the quantity to be made largest or smallest.
  2. Use the given condition (constraint) to write it as a function of one variable, and state the allowed interval.
  3. Find the critical values.
  4. Test them (second derivative test, or compare with the endpoint values).
  5. Answer the question that was asked, with units.
Example 52 (book)
Find two non-negative numbers whose sum is 15 such that the product of one with the square of the other is a maximum.
  1. Let the numbers be x\ge0 and y\ge0 with x+y=15, and let p=xy^{2}. Substitute y=15-x.
    p(x)=x(15-x)^{2},\qquad0\le x\le15
  2. Differentiate (product rule) and factorise.
    p'(x)=(15-x)^{2}+x\cdot2(15-x)(-1)=(15-x)(15-3x)
  3. Critical value inside the interval: x=5 (the other root, x=15, is an endpoint).
  4. Compare with the endpoints.
    p(0)=0,\qquad p(15)=0,\qquad p(5)=5(10)^{2}=500
The product is largest when x = 5.
x=5,\ y=10,\ \text{maximum product }500
Example 53 (book, corrected)
A rectangular plot of land of area 1500\,\mathrm{m^{2}} is to be fenced and divided into two equal portions by an additional fence parallel to two sides. Find the dimensions that require the least amount of fencing.
Two sides of length x and three fences of length y.
  1. Total fence and the area condition.
    L=2x+3y,\qquad xy=1500\ \Longrightarrow\ y=\dfrac{1500}{x}
  2. Write L in terms of x only.
    L(x)=2x+\dfrac{4500}{x},\qquad x>0
  3. Critical value.
    L'(x)=2-\dfrac{4500}{x^{2}}=0\ \Longrightarrow\ x^{2}=2250\ \Longrightarrow\ x=15\sqrt{10}\approx47.4\,\mathrm{m}
  4. Second derivative test.
    L''(x)=\dfrac{9000}{x^{3}}>0\ \text{for }x>0\ \Longrightarrow\ \text{minimum}
  5. The other side and the least fencing.
    \begin{aligned}y&=\dfrac{1500}{15\sqrt{10}}=10\sqrt{10}\approx31.6\,\mathrm{m}\\ L&=2(15\sqrt{10})+3(10\sqrt{10})\\ &=60\sqrt{10}\approx189.7\,\mathrm{m}\end{aligned}
15\sqrt{10}\,\mathrm{m}\times10\sqrt{10}\,\mathrm{m}\ (\approx47.4\,\mathrm{m}\times31.6\,\mathrm{m})

Price growth and inflation

With a continuous annual inflation rate r, the price level at time t years is P(t)=P_0e^{rt}, where P_0 is the initial price. The derivative gives how fast prices are rising at time t:

P(t)=P_0e^{rt}\qquad\Longrightarrow\qquad\dfrac{dP}{dt}=P_0re^{rt}=rP(t)
Example 54 (book)
The price of a product is P(t)=200e^{0.03t} (t in years). Find the rate at which the price is increasing after (a) 0 years (b) 5 years (c) 10 years.
  1. Differentiate.
    \dfrac{dP}{dt}=200(0.03)e^{0.03t}=6e^{0.03t}
  2. (a) t=0.
    6e^{0}=6\ \text{units per year}
  3. (b) t=5.
    6e^{0.15}=6(1.1618)=6.97\ \text{units per year}
  4. (c) t=10.
    6e^{0.3}=6(1.3499)=8.10\ \text{units per year}

Inflation over a period. The percentage rise in price from time t_1 to t_2 is \dfrac{P(t_2)-P(t_1)}{P(t_1)}\times100\%. For the model P_0e^{rt} this is \left(e^{r(t_2-t_1)}-1\right)\times100\%; with r=0.03 over 5 years it is (e^{0.15}-1)\times100\%\approx16.2\%. For actual prices, if a basket of goods cost Rs 2000 in 2020 and Rs 2600 in 2025, inflation over the period is \dfrac{600}{2000}\times100\%=30\%.

Straight-line models: marginal cost and revenue

When a quantity changes at a constant rate its graph is a straight line, and the derivative is the constant slope. In economics, the marginal revenue R'(x) and marginal cost C'(x) are the extra revenue and extra cost from one more unit.

Example 55 (book)
A company's revenue from selling x units is R(x)=50x and the total cost is C(x)=30x+200. (a) Find the marginal revenue and marginal cost. (b) Find the break-even point. (c) Interpret the slopes.
  1. (a) Differentiate.
    R'(x)=50,\qquad C'(x)=30
  2. (b) At break-even, revenue equals cost.
    50x=30x+200\ \Longrightarrow\ x=10
  3. (c) Revenue grows by Rs 50 per unit and cost by only Rs 30 per unit, so each unit sold after the break-even point adds Rs 20 of profit. The 200 in C(x) is the fixed cost.
R'=50,\ C'=30,\ \text{break-even at }10\ \text{units}

Straight-line depreciation

A machine loses value as it ages. In the straight-line method it loses the same amount D every year: D=\dfrac{C-S}{n}, where C is the cost, S the salvage (scrap) value and n the useful life in years. Its book value after t years is V(t)=C-Dt, and \dfrac{dV}{dt}=-D: the value falls at a constant rate.

Example: straight-line depreciation
A machine costs Rs 500 000 and has a salvage value of Rs 50 000 after 10 years. Find the annual depreciation, the value after 4 years, and the rate of change of its value.
  1. Annual depreciation.
    D=\dfrac{500\,000-50\,000}{10}=45\,000\ \text{rupees per year}
  2. Value after t years, then t=4.
    V(t)=500\,000-45\,000t,\qquad V(4)=320\,000\ \text{rupees}
  3. Rate of change.
    \dfrac{dV}{dt}=-45\,000\ \text{rupees per year}

Formula sheet

Existence of a limit
\lim_{x\to a}f(x)=L\iff\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=L
a^-, a^+: approach from the left and from the right; L: a finite number
Limit theorems
\begin{aligned}\lim(f\pm g)&=L_1\pm L_2\\ \lim(fg)&=L_1L_2\\ \lim\dfrac fg&=\dfrac{L_1}{L_2}\ (L_2\neq0)\\ \lim f^{n}&=L_1^{\,n}\end{aligned}
L_1=\lim_{x\to a}f(x), L_2=\lim_{x\to a}g(x)
Limit of a difference of powers
\lim_{x\to a}\dfrac{x^{n}-a^{n}}{x-a}=na^{n-1}
n: a rational number
Trigonometric limits
\lim_{x\to0}\dfrac{\sin x}{x}=1,\qquad\lim_{x\to0}\dfrac{1-\cos x}{x}=0,\qquad\lim_{x\to0}\dfrac{\tan x}{x}=1
x in radians
Limits defining e
\lim_{n\to\infty}\left(1+\dfrac1n\right)^{n}=e,\qquad\lim_{x\to0}(1+x)^{1/x}=e,\qquad\lim_{x\to\infty}\left(1+\dfrac ax\right)^{bx}=e^{ab}
e\approx2.71828
Exponential limits
\lim_{x\to0}\dfrac{e^{x}-1}{x}=1,\qquad\lim_{x\to0}\dfrac{a^{x}-1}{x}=\ln a
a>0
Continuity at a number
f\ \text{continuous at }a\iff\lim_{x\to a}f(x)=f(a)
requires f(a) defined and the limit existing
Slope of a secant
m_{\sec}=\dfrac{\Delta y}{\Delta x}=\dfrac{f(a+\Delta x)-f(a)}{\Delta x}
\Delta x: change in x; \Delta y: change in y (also the average rate of change)
Slope of a tangent
m_{\tan}=\lim_{\Delta x\to0}\dfrac{f(a+\Delta x)-f(a)}{\Delta x}=f'(a)
slope of the curve at (a,f(a)) (instantaneous rate of change)
Equation of the tangent
y-f(a)=f'(a)(x-a)
point (a,f(a)), slope f'(a)
Instantaneous velocity
v(t_1)=\lim_{\Delta t\to0}\dfrac{f(t_1+\Delta t)-f(t_1)}{\Delta t}
s=f(t): position at time t
Unit: \mathrm{m\,s^{-1}}
Derivative (first principles)
f'(x)=\dfrac{dy}{dx}=\lim_{\Delta x\to0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}
y=f(x)
Power rule
\dfrac{d}{dx}x^{n}=nx^{n-1}
n: any real number
Constant, multiple and sum rules
\dfrac{d}{dx}c=0,\quad\dfrac{d}{dx}[cf]=cf',\quad\dfrac{d}{dx}[f\pm g]=f'\pm g'
c: a constant
Product rule
\dfrac{d}{dx}[fg]=fg'+gf'
f, g: differentiable functions of x
Quotient rule
\dfrac{d}{dx}\left[\dfrac fg\right]=\dfrac{gf'-fg'}{g^{2}}
g(x)\neq0
Power rule for functions
\dfrac{d}{dx}[g(x)]^{n}=n[g(x)]^{n-1}g'(x)
g: differentiable
Chain rule
\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}=f'(g(x))\,g'(x)
y=f(u), u=g(x)
Derivatives of sin and cos
\dfrac{d}{dx}\sin x=\cos x,\qquad\dfrac{d}{dx}\cos x=-\sin x
x in radians
Derivatives of tan and cot
\dfrac{d}{dx}\tan x=\sec^{2}x,\qquad\dfrac{d}{dx}\cot x=-\operatorname{cosec}^{2}x
x in radians
Derivatives of sec and cosec
\dfrac{d}{dx}\sec x=\sec x\tan x,\qquad\dfrac{d}{dx}\operatorname{cosec} x=-\operatorname{cosec} x\cot x
x in radians
Derivatives of inverse sine and cosine
\dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^{2}}},\qquad\dfrac{d}{dx}\cos^{-1}x=-\dfrac{1}{\sqrt{1-x^{2}}}
-1<x<1
Derivatives of inverse tangent and cotangent
\dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^{2}},\qquad\dfrac{d}{dx}\cot^{-1}x=-\dfrac{1}{1+x^{2}}
all real x
Derivatives of inverse secant and cosecant
\dfrac{d}{dx}\sec^{-1}x=\dfrac{1}{x\sqrt{x^{2}-1}},\qquad\dfrac{d}{dx}\operatorname{cosec}^{-1}x=-\dfrac{1}{x\sqrt{x^{2}-1}}
|x|>1 (book's choice of range)
Exponential derivatives
\dfrac{d}{dx}e^{x}=e^{x},\qquad\dfrac{d}{dx}a^{x}=a^{x}\ln a,\qquad\dfrac{d}{dx}e^{u}=e^{u}u'
a>0, a\neq1; u: a function of x
Logarithmic derivatives
\dfrac{d}{dx}\ln x=\dfrac1x,\qquad\dfrac{d}{dx}\log_a x=\dfrac{1}{x\ln a},\qquad\dfrac{d}{dx}\ln u=\dfrac{u'}{u}
x>0, u>0
Parametric derivative
\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}
x=f(t), y=g(t), \dfrac{dx}{dt}\neq0
Differential
dy=f'(x)\,dx,\qquad\Delta y\approx dy
dx=\Delta x: change in x; dy: change along the tangent
Linear approximation
f(x+\Delta x)\approx f(x)+f'(x)\,dx
x: a nearby number where f(x) is known
Relative and percentage error
\begin{aligned}\text{relative error}&=\dfrac{dy}{y}\\ \text{percentage error}&=\dfrac{dy}{y}\times100\%\\ y=kx^{n}\Rightarrow\dfrac{dy}{y}&=n\dfrac{dx}{x}\end{aligned}
dx: error in the measured quantity x
Higher derivatives
f''(x)=\dfrac{d^{2}y}{dx^{2}}=\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right),\qquad\dfrac{d^{n}y}{dx^{n}}=\dfrac{d}{dx}\left(\dfrac{d^{n-1}y}{dx^{n-1}}\right)
n: a positive integer
Increasing and decreasing test
f'(x)>0\ \text{on }(a,b)\Rightarrow f\ \text{increasing};\qquad f'(x)<0\Rightarrow f\ \text{decreasing}
f differentiable on (a,b)
Critical value
f'(c)=0\ \text{or}\ f'(c)\ \text{does not exist},\quad c\in D_f
D_f: domain of f
Test for concavity
f''(x)>0\Rightarrow\text{concave upward};\qquad f''(x)<0\Rightarrow\text{concave downward}
on an interval where f'' exists
Second derivative test
f'(c)=0:\quad f''(c)>0\Rightarrow\text{rel. min},\qquad f''(c)<0\Rightarrow\text{rel. max}
if f''(c)=0 the test fails
Related rates (square and sphere)
A=x^{2}\Rightarrow\dfrac{dA}{dt}=2x\dfrac{dx}{dt},\qquad V=\tfrac43\pi r^{3}\Rightarrow\dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}
t: time
Price growth (inflation) model
P(t)=P_0e^{rt},\qquad\dfrac{dP}{dt}=rP
P_0: initial price; r: continuous annual inflation rate; t: time in years
Straight-line depreciation
D=\dfrac{C-S}{n},\qquad V(t)=C-Dt,\qquad\dfrac{dV}{dt}=-D
C: cost; S: salvage value; n: useful life in years; V: book value
Unit: currency per year for D
Marginal revenue and cost
MR=R'(x),\qquad MC=C'(x);\qquad\text{break-even: }R(x)=C(x)
R: revenue, C: cost, x: number of units

Key terms

Limit
The number L that f(x) approaches as x approaches a from both sides, written \lim_{x\to a}f(x)=L.
Left-hand limit
The value f(x) approaches as x\to a through values less than a, written \lim_{x\to a^-}f(x).
Right-hand limit
The value f(x) approaches as x\to a through values greater than a, written \lim_{x\to a^+}f(x).
Indeterminate form
An expression such as \dfrac00 obtained by substitution; it gives no information until the function is simplified.
Continuous at a number
f is continuous at a if f(a) is defined, \lim_{x\to a}f(x) exists and \lim_{x\to a}f(x)=f(a).
Discontinuous
Not continuous: at least one of the three conditions of continuity fails at the number.
Continuous on an interval
Continuous at every number of an open interval; on a closed interval [a,b] also continuous from the right at a and from the left at b.
Removable discontinuity
A discontinuity where the limit exists; defining f(a) equal to the limit removes it.
Jump discontinuity
A discontinuity where the left-hand and right-hand limits exist but are different.
Continuous extension
The function obtained by defining (or redefining) f(a) as \lim_{x\to a}f(x) at a removable discontinuity.
Intermediate value theorem
A function continuous on [a,b] takes every value between f(a) and f(b); so a sign change means a root.
Secant line
A line through two points of a curve; its slope \dfrac{\Delta y}{\Delta x} is the average rate of change.
Tangent line
The line through (a,f(a)) with slope \lim_{\Delta x\to0}\dfrac{\Delta y}{\Delta x}, whenever this limit exists.
Average rate of change
\dfrac{f(a+\Delta x)-f(a)}{\Delta x}, the change in f per unit change in x over an interval.
Instantaneous rate of change
The limit of the average rate of change as \Delta x\to0; the derivative at a point.
Instantaneous velocity
v(t)=\lim_{\Delta t\to0}\dfrac{\Delta s}{\Delta t}, the rate of change of position at an instant.
Derivative
f'(x)=\lim_{\Delta x\to0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}, the slope of the curve and the rate of change of f.
Differentiable
A function is differentiable at x if its derivative exists (as a finite limit) there.
Differentiation
The process of finding a derivative; “from first principles” means directly from the limit definition.
Chain rule
\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}: the derivative of a composite function.
Explicit function
A function written as y=f(x), with y alone on one side.
Implicit function
A relation in which x and y are mixed, such as 5y^{2}+\sin y=x^{2}.
Implicit differentiation
Differentiating both sides of an equation in x and y with respect to x, treating y as a function of x.
Parametric equations
Equations x=f(t), y=g(t) that give both coordinates in terms of a parameter t.
Logarithmic differentiation
Taking logarithms before differentiating, used for functions such as x^{x}.
Differential
dx=\Delta x is the differential of x; dy=f'(x)\,dx is the differential of y.
Absolute error
The approximate error dy in a calculated quantity caused by an error dx in a measurement.
Relative error
The error as a fraction of the value, \dfrac{dy}{y}.
Percentage error
The relative error times 100, \dfrac{dy}{y}\times100\%.
Higher order derivative
A derivative of a derivative: f'', f''', f^{(4)}, and so on.
Increasing function
f is increasing on I if x_1<x_2 in I implies f(x_1)<f(x_2).
Decreasing function
f is decreasing on I if x_1<x_2 in I implies f(x_1)>f(x_2).
Absolute maximum
f(c) with f(x)\le f(c) for every x in the domain; also called the global maximum.
Absolute minimum
f(c) with f(x)\ge f(c) for every x in the domain; also called the global minimum.
Extreme value theorem
A function continuous on a closed interval [a,b] has an absolute maximum and an absolute minimum on it.
Relative maximum
f(c) with f(x)\le f(c) for every x in some open interval containing c (local maximum).
Relative minimum
f(c) with f(x)\ge f(c) for every x in some open interval containing c (local minimum).
Critical value
A number c in the domain of f where f'(c)=0 or f'(c) does not exist; (c,f(c)) is a critical point.
Stationary point
A point where f'(c)=0, so the tangent is horizontal.
Concave upward
The graph bends up like a cup (“holds water”); f''>0.
Concave downward
The graph bends down like a cap (“spills water”); f''<0.
Point of inflection
A point on a continuous graph where the concavity changes from upward to downward or the reverse.
Optimisation
Finding the largest or smallest value of a quantity, usually by locating critical values.
Marginal cost
C'(x), the approximate cost of producing one more unit.
Marginal revenue
R'(x), the approximate revenue from selling one more unit.
Break-even point
The number of units at which revenue equals cost, R(x)=C(x).
Inflation rate
The rate at which the price level rises; with P=P_0e^{rt}, r is the continuous annual rate.
Straight-line depreciation
A method in which an asset loses the same value D=\dfrac{C-S}{n} every year.

Exercise 2.1p. 46–47

Limits from graphs, by direct substitution, by factorising or rationalising a \dfrac00 form, and with the standard limit \lim_{x\to0}\dfrac{\sin x}{x}=1.

1
Use a graph to find the given limit, if it exists.
Solution
(a) \lim_{x\to5}\sqrt{x-1}
  1. Domain. \sqrt{x-1} is defined for x\ge1, so values of x on both sides of 5 are allowed.
  2. Graph. y=\sqrt{x-1} is the curve y=\sqrt x moved 1 unit right. It has no break near x=5.
  3. Read the limit. As x\to5 from either side, the curve heads to the height \sqrt{5-1}=2. A table agrees: 4.9\to1.975, 4.99\to1.9975, 5.01\to2.0025, 5.1\to2.0248.
    \lim_{x\to5^-}\sqrt{x-1}=\lim_{x\to5^+}\sqrt{x-1}=2
\lim_{x\to5}\sqrt{x-1}=2
(b) \lim_{x\to1}\dfrac{x^{2}-1}{x-1}
  1. The function is not defined at x=1 (\dfrac00). For x\neq1 cancel the factor x-1.
    \dfrac{x^{2}-1}{x-1}=\dfrac{(x-1)(x+1)}{x-1}=x+1,\quad x\neq1
  2. Graph. The line y=x+1 with a hole at (1,2).
  3. Both sides of the line approach the height of the hole.
    \lim_{x\to1}\dfrac{x^{2}-1}{x-1}=1+1=2
2
(c) \lim_{x\to0}\dfrac{x^{2}-3x}{x}
  1. Not defined at x=0. For x\neq0 divide by x.
    \dfrac{x^{2}-3x}{x}=\dfrac{x(x-3)}{x}=x-3,\quad x\neq0
  2. Graph. The line y=x-3 with a hole at (0,-3).
  3. Read the height of the hole.
    \lim_{x\to0}\dfrac{x^{2}-3x}{x}=0-3=-3
-3
(d) \lim_{x\to0}\dfrac{|x|}{x}
  1. For x>0, |x|=x, so \dfrac{|x|}{x}=1. For x<0, |x|=-x, so \dfrac{|x|}{x}=-1. The function is not defined at 0.
    \dfrac{|x|}{x}=\begin{cases}-1, & x<0\\ 1, & x>0\end{cases}
  2. Graph. Two horizontal half-lines with hollow ends at (0,-1) and (0,1).
  3. The one-sided limits are different.
    \lim_{x\to0^-}\dfrac{|x|}{x}=-1\neq1=\lim_{x\to0^+}\dfrac{|x|}{x}
The graph jumps from -1 to 1 at x = 0.
\text{does not exist}
(e) \lim_{x\to2}f(x), where f(x)=\begin{cases}x, & x<2\\ x+1, & x\ge2\end{cases}
  1. Left of 2 the rule is f(x)=x.
    \lim_{x\to2^-}f(x)=\lim_{x\to2^-}x=2
  2. Right of 2 the rule is f(x)=x+1.
    \lim_{x\to2^+}f(x)=\lim_{x\to2^+}(x+1)=3
  3. The graph jumps from height 2 to height 3 at x=2, so the one-sided limits differ.
    2\neq3
\text{does not exist}
(f) \lim_{x\to0}f(x), where f(x)=\begin{cases}x^{2}, & x<0\\ 2, & x=0\\ \sqrt{x}-1, & x>0\end{cases}
  1. Left of 0 the rule is x^{2}.
    \lim_{x\to0^-}x^{2}=0
  2. Right of 0 the rule is \sqrt x-1.
    \lim_{x\to0^+}(\sqrt x-1)=0-1=-1
  3. The value f(0)=2 is an isolated point and plays no part in the limit. The one-sided limits differ.
    0\neq-1
The left piece ends at (0, 0), the right piece starts at (0, -1), and f(0) = 2 is a separate dot.
\text{does not exist}
(g) \lim_{x\to0}\dfrac{1-\cos x}{x^{2}}
  1. The function is not defined at 0 (\dfrac00) but it is even, so both sides behave the same. A table (radians):
    \begin{array}{c|cccc} x & \pm0.5 & \pm0.1 & \pm0.01 & \pm0.001\\ \hline \dfrac{1-\cos x}{x^{2}} & 0.4897 & 0.49958 & 0.4999958 & 0.5000000\end{array}
  2. Graph. A hump whose top has a hole at (0,\tfrac12).
  3. Check by algebra. Use 1-\cos x=2\sin^{2}\dfrac x2 and \lim_{u\to0}\dfrac{\sin u}{u}=1 with u=\dfrac x2.
    \dfrac{1-\cos x}{x^{2}}=\dfrac{2\sin^{2}(x/2)}{x^{2}}=\dfrac12\left(\dfrac{\sin(x/2)}{x/2}\right)^{2}\to\dfrac12(1)^{2}=\dfrac12
\dfrac12
2
Use the given graph to find each limit (x\to1), if it exists.
Solution
(a) For the graph below.
Redrawn from the book: both lines end at a hollow dot at (1,3); a separate dot on the x-axis at x=1.
  1. From the left: the rising line approaches the hollow dot at height 3.
    \lim_{x\to1^-}f(x)=3
  2. From the right: the falling line also approaches the hollow dot at height 3.
    \lim_{x\to1^+}f(x)=3
  3. The one-sided limits agree, so the limit exists. The dot on the x-axis shows that f(1)=0; the value of f at 1 does not affect the limit.
    \lim_{x\to1}f(x)=3\quad(\text{although }f(1)=0)
\lim_{x\to1}f(x)=3
(b) For the graph below.
Redrawn from the book: the left piece ends at a hollow dot at (1,2); the right piece plunges downward as x\to1^{+}.
  1. From the left: the curve approaches the hollow dot at height 2.
    \lim_{x\to1^-}f(x)=2
  2. From the right: the curve falls without bound (the arrow points downward along x=1), so there is no finite right-hand limit.
    \lim_{x\to1^+}f(x)=-\infty
  3. The right-hand limit does not exist as a number, and in any case it is not equal to the left-hand limit.
\lim_{x\to1}f(x)\ \text{does not exist}

Note. The book's graphs are redrawn from the printed figure; only their shape near x = 1 matters.

3
Evaluate \lim_{x\to-2}\dfrac{x^{3}+8}{x+2}.
Solution
  1. Substituting x=-2 gives \dfrac{-8+8}{0}=\dfrac00, so simplify first. Factorise the sum of cubes a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2}).
    x^{3}+8=(x+2)(x^{2}-2x+4)
  2. Cancel x+2 (allowed because x\neq-2 as x\to-2).
    \lim_{x\to-2}\dfrac{(x+2)(x^{2}-2x+4)}{x+2}=\lim_{x\to-2}(x^{2}-2x+4)
  3. Substitute.
    =(-2)^{2}-2(-2)+4=4+4+4=12
12
4
Evaluate \lim_{x\to4}\dfrac{\sqrt{x}-2}{x-4}.
Solution
  1. Substitution gives \dfrac{2-2}{0}=\dfrac00. Write x-4 as a difference of squares: x-4=(\sqrt x-2)(\sqrt x+2).
    \dfrac{\sqrt x-2}{x-4}=\dfrac{\sqrt x-2}{(\sqrt x-2)(\sqrt x+2)}=\dfrac{1}{\sqrt x+2},\quad x\neq4
  2. Substitute x=4.
    \lim_{x\to4}\dfrac{1}{\sqrt x+2}=\dfrac{1}{2+2}=\dfrac14
\dfrac14
5
Evaluate \lim_{x\to7}\dfrac{x^{2}-21}{x+2}.
Solution
  1. The denominator tends to 7+2=9\neq0, so the quotient theorem applies: substitute directly.
    \lim_{x\to7}\dfrac{x^{2}-21}{x+2}=\dfrac{49-21}{7+2}=\dfrac{28}{9}
\dfrac{28}{9}
6
Evaluate \lim_{x\to0}\dfrac{x^{2}-6x}{x^{2}-7x+6}.
Solution
  1. The denominator tends to 0-0+6=6\neq0, so substitute directly.
    \lim_{x\to0}\dfrac{x^{2}-6x}{x^{2}-7x+6}=\dfrac{0-0}{6}=0
  2. Check by factorising (valid for x\neq6): the fraction is \dfrac{x(x-6)}{(x-1)(x-6)}=\dfrac{x}{x-1}, which also gives \dfrac{0}{-1}=0.
0
7
Evaluate \lim_{y\to1}\dfrac{y^{3}-1}{y-1}.
Solution
  1. Substitution gives \dfrac00. Factorise the difference of cubes a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2}).
    y^{3}-1=(y-1)(y^{2}+y+1)
  2. Cancel y-1 (y\neq1) and substitute.
    \lim_{y\to1}(y^{2}+y+1)=1+1+1=3
3
8
Evaluate \lim_{x\to3^{+}}\dfrac{(x+3)^{2}}{\sqrt{x-3}}.
Solution
  1. Only x>3 is allowed (\sqrt{x-3} needs x\ge3 and the denominator must not be 0). The numerator tends to (3+3)^{2}=36.
  2. The denominator \sqrt{x-3} tends to 0 through positive values. The form is \dfrac{36}{0^{+}}, so the fraction becomes larger than any number.
    \begin{array}{c|ccc} x & 3.1 & 3.01 & 3.001\\ \hline \dfrac{(x+3)^{2}}{\sqrt{x-3}} & 117.7 & 361.2 & 1138.8\end{array}
  3. So the fraction increases without bound.
    \lim_{x\to3^{+}}\dfrac{(x+3)^{2}}{\sqrt{x-3}}=+\infty
The graph rises steeply along the asymptote x = 3.
+\infty\ (\text{the limit does not exist as a real number})
9
Evaluate \lim_{x\to2}(x-4)^{4}(x^{2}-3)^{10}.
Solution
  1. A product of powers of polynomials: use the product and power theorems, that is, substitute directly.
    \lim_{x\to2}(x-4)^{4}(x^{2}-3)^{10}=(2-4)^{4}(4-3)^{10}
  2. Evaluate.
    =(-2)^{4}(1)^{10}=16
16
10
Evaluate \lim_{x\to0}\left(x-\dfrac{1}{x-2}\right).
Solution
  1. The denominator x-2\to-2\neq0, so the difference theorem applies and we substitute.
    \lim_{x\to0}\left(x-\dfrac{1}{x-2}\right)=0-\dfrac{1}{0-2}=\dfrac12
\dfrac12
11
Evaluate \lim_{x\to-3}\dfrac{2x+6}{4x^{2}-36}.
Solution
  1. Substitution gives \dfrac{-6+6}{36-36}=\dfrac00. Factorise the top and the bottom.
    \dfrac{2x+6}{4x^{2}-36}=\dfrac{2(x+3)}{4(x-3)(x+3)}=\dfrac{1}{2(x-3)},\quad x\neq-3
  2. Substitute x=-3.
    \lim_{x\to-3}\dfrac{1}{2(x-3)}=\dfrac{1}{2(-6)}=-\dfrac{1}{12}
-\dfrac{1}{12}
12
Evaluate \lim_{x\to0}\dfrac{\tan x}{x}.
Solution
  1. Write \tan x=\dfrac{\sin x}{\cos x} to bring in the standard limit.
    \dfrac{\tan x}{x}=\dfrac{\sin x}{x}\cdot\dfrac{1}{\cos x}
  2. Product theorem with \lim_{x\to0}\dfrac{\sin x}{x}=1 and \cos0=1.
    \lim_{x\to0}\dfrac{\tan x}{x}=1\cdot\dfrac11=1
1
13
Evaluate \lim_{x\to0}\dfrac{x}{\sin3x}.
Solution
  1. Make the angle and the denominator match: multiply and divide by 3.
    \dfrac{x}{\sin3x}=\dfrac13\cdot\dfrac{3x}{\sin3x}
  2. As x\to0, u=3x\to0 and \dfrac{u}{\sin u}\to1.
    \lim_{x\to0}\dfrac{x}{\sin3x}=\dfrac13\cdot1=\dfrac13
\dfrac13

Exercise 2.2p. 50–51

Test the three conditions of continuity: f(a) is defined, \lim_{x\to a}f(x) exists, and the two are equal. For a piecewise function, test every joining point; for an interval, also check the included endpoints.

1
Determine the numbers (if any) at which the given function is discontinuous: f(x)=x^{2}-5x+6.
Solution
  1. f is a polynomial. For every real a, \lim_{x\to a}f(x)=f(a) by direct substitution, so all three conditions of continuity hold at every number.
    \lim_{x\to a}(x^{2}-5x+6)=a^{2}-5a+6=f(a)
\text{no discontinuities: continuous for all real }x
2
Determine the numbers (if any) at which the given function is discontinuous: f(x)=\dfrac{2x}{x^{2}+5}.
Solution
  1. A rational function is discontinuous only where its denominator is zero. Here the denominator is never zero:
    x^{2}+5\ge5>0\quad\text{for all real }x
  2. So f(a) is defined and \lim_{x\to a}f(x)=\dfrac{2a}{a^{2}+5}=f(a) for every a.
\text{no discontinuities}
3
Determine the numbers (if any) at which the given function is discontinuous: f(x)=\dfrac{1}{x^{2}-9x+8}.
Solution
  1. Find where the denominator is zero.
    x^{2}-9x+8=(x-1)(x-8)=0\ \Longrightarrow\ x=1\ \text{or}\ x=8
  2. At x=1 and x=8, f is not defined, so condition (i) fails. Near these numbers the values grow without bound (infinite discontinuities). Everywhere else f is a rational function with a non-zero denominator, so it is continuous.
Vertical asymptotes at x = 1 and x = 8.
x=1\ \text{and}\ x=8
4
Determine the numbers (if any) at which the given function is discontinuous: f(x)=\dfrac{x^{2}-1}{x^{4}-1}.
Solution
  1. Find where the denominator is zero.
    x^{4}-1=(x^{2}-1)(x^{2}+1)=0\ \Longrightarrow\ x^{2}=1\ \Longrightarrow\ x=\pm1
  2. f(-1) and f(1) are not defined, so f is discontinuous at x=\pm1. (The factor x^{2}+1 is never zero.)
  3. Both discontinuities are removable: for x\neq\pm1, f(x)=\dfrac{1}{x^{2}+1}, so the limit at each point is \dfrac12.
    \lim_{x\to\pm1}f(x)=\dfrac{1}{1+1}=\dfrac12
Two holes, at x = -1 and x = 1.
x=-1\ \text{and}\ x=1
5
Determine the numbers (if any) at which the given function is discontinuous: f(x)=\dfrac{x-1}{\sin2x}.
Solution
  1. Numerator and denominator are continuous everywhere, so f can fail only where \sin2x=0.
    \sin2x=0\iff2x=n\pi\iff x=\dfrac{n\pi}{2},\quad n\in\mathbb{Z}
  2. At each such number f is not defined (the numerator x-1 is not zero there, so these are infinite discontinuities).
x=\dfrac{n\pi}{2},\ n\in\mathbb{Z}\quad(\dots,-\pi,-\tfrac{\pi}{2},0,\tfrac{\pi}{2},\pi,\dots)
6
Determine the numbers (if any) at which the given function is discontinuous: f(x)=\begin{cases}x, & x<0\\ x^{2}, & 0\le x\le2\\ x, & x\ge2\end{cases}.
Solution
  1. Each piece is a polynomial, so f is continuous inside each interval. Only the joining points x=0 and x=2 need testing.
  2. **At x=0.** f(0) comes from the middle piece. The left limit uses x, the right limit uses x^{2}.
    f(0)=0,\qquad\lim_{x\to0^-}x=0,\qquad\lim_{x\to0^+}x^{2}=0\ \Longrightarrow\ \text{continuous at }0
  3. **At x=2.** The left limit uses x^{2}, the right limit uses x.
    \lim_{x\to2^-}x^{2}=4,\qquad\lim_{x\to2^+}x=2
  4. The one-sided limits differ, so \lim_{x\to2}f(x) does not exist and f is discontinuous at 2 (a jump).
The graph is unbroken at x = 0 but jumps from 4 down to 2 at x = 2.
x=2\ \text{only}

Note. The book prints the conditions as 0\le x\le2 and x\ge2, which overlap at x=2 with different values (4 and 2). The third piece is taken as x>2, so f(2)=4. With either reading f is discontinuous at x=2.

7
Determine the numbers (if any) at which the given function is discontinuous: f(x)=\begin{cases}\dfrac{\sin x}{x}, & x\neq0\\ \dfrac12, & x=0\end{cases}.
Solution
  1. For x\neq0, f is a quotient of continuous functions with a non-zero denominator, so it is continuous there. Test x=0.
  2. Condition (i): f(0) is defined.
    f(0)=\dfrac12
  3. Condition (ii): the standard limit exists.
    \lim_{x\to0}\dfrac{\sin x}{x}=1
  4. Condition (iii) fails, so f is discontinuous at 0 (removable: redefining f(0)=1 would make it continuous).
    \lim_{x\to0}f(x)=1\neq\dfrac12=f(0)
x=0\ \text{only}
8
Determine the numbers (if any) at which the given function is discontinuous: f(x)=\begin{cases}\dfrac{x^{2}-36}{x-6}, & x\neq6\\ 12, & x=6\end{cases}.
Solution
  1. For x\neq6, cancel the factor x-6: f(x)=x+6, a polynomial, so f is continuous at every x\neq6.
    \dfrac{x^{2}-36}{x-6}=\dfrac{(x-6)(x+6)}{x-6}=x+6
  2. At x=6: the value and the limit are equal.
    \lim_{x\to6}f(x)=6+6=12=f(6)
  3. All three conditions hold at 6 as well.
\text{no discontinuities: continuous for all real }x
9
Determine whether f(x)=x^{2}+1 is continuous in the indicated intervals.
Solution
(a) [-1,3]
  1. f is a polynomial, so it is continuous at every number in (-1,3).
  2. At the endpoints the one-sided limits equal the values.
    \lim_{x\to-1^+}(x^{2}+1)=2=f(-1),\qquad\lim_{x\to3^-}(x^{2}+1)=10=f(3)
\text{continuous on }[-1,3]
(b) [3,\infty)
  1. Continuous at every x>3 (polynomial), and at the endpoint:
    \lim_{x\to3^+}(x^{2}+1)=10=f(3)
\text{continuous on }[3,\infty)
10
Determine whether f(x)=\dfrac1x is continuous in the indicated intervals.
Solution
1/x breaks only at x = 0, which lies inside (-3, 3) but outside (0, 10].
(a) (-3,3)
  1. The interval contains 0, and f(0)=\dfrac10 is not defined.
  2. So f is discontinuous at a number of the interval (the graph has a vertical asymptote at x=0).
\text{not continuous on }(-3,3)
(b) (0,10]
  1. For 0<x\le10, x\neq0, so f is defined and continuous on (0,10). The open end at 0 is not included, so the blow-up there does not matter.
  2. At the included endpoint 10:
    \lim_{x\to10^-}\dfrac1x=\dfrac1{10}=f(10)
\text{continuous on }(0,10]
11
Determine whether f(x)=\dfrac{1}{\sqrt x} is continuous in the indicated intervals.
Solution
(a) [1,4)
  1. f is defined and continuous for every x>0 (quotient of continuous functions, denominator \sqrt x>0). The interval [1,4) lies inside x>0.
  2. At the included endpoint 1:
    \lim_{x\to1^+}\dfrac1{\sqrt x}=1=f(1)
\text{continuous on }[1,4)
(b) [1,9]
  1. [1,9] also lies inside x>0, so f is continuous on (1,9).
  2. Both endpoints check.
    \lim_{x\to1^+}\dfrac1{\sqrt x}=1=f(1),\qquad\lim_{x\to9^-}\dfrac1{\sqrt x}=\dfrac13=f(9)
\text{continuous on }[1,9]
12
Determine whether f(x)=\sqrt{x^{2}-9} is continuous in the indicated intervals.
Solution
No graph at all between -3 and 3.
(a) [-3,3]
  1. \sqrt{x^{2}-9} is real only when x^{2}\ge9, that is x\le-3 or x\ge3.
  2. For -3<x<3 (for example x=0, giving \sqrt{-9}) f is not defined, so condition (i) fails at those numbers.
\text{not continuous on }[-3,3]
(b) [3,\infty)
  1. For x>3, x^{2}-9>0 and f is a composition of continuous functions, so it is continuous.
  2. At the endpoint 3:
    \lim_{x\to3^+}\sqrt{x^{2}-9}=0=f(3)
\text{continuous on }[3,\infty)
13
Determine whether f(x)=\dfrac{x}{x^{3}+8} is continuous in the indicated intervals.
Solution
(a) [-4,-3]
  1. Find where the denominator is zero. The quadratic factor has discriminant 4-16<0, so it has no real roots.
    x^{3}+8=(x+2)(x^{2}-2x+4)=0\ \Longrightarrow\ x=-2
  2. -2 is not in [-4,-3], so f is a rational function with a non-zero denominator on the whole interval (including the endpoints).
\text{continuous on }[-4,-3]
(b) [-10,10]
  1. -2 lies in [-10,10] and f(-2) is not defined.
\text{not continuous on }[-10,10]
14
Determine whether f(x)=\sin\dfrac1x is continuous in the indicated intervals.
Solution
The graph oscillates wildly only near x=0; on intervals away from 0 it is unbroken.
(a) \left[\dfrac1\pi,5\right)
  1. \dfrac1x is continuous for x\neq0 and \sin is continuous everywhere, so the composition \sin\dfrac1x is continuous for every x\neq0.
  2. The interval \left[\dfrac1\pi,5\right) does not contain 0 (and at the endpoint \dfrac1\pi, \lim_{x\to1/\pi^+}\sin\dfrac1x=\sin\pi=0=f\!\left(\tfrac1\pi\right)).
\text{continuous on }\left[\tfrac1\pi,5\right)
(b) \left[\dfrac\pi2,\dfrac{3\pi}2\right]
  1. This interval also avoids 0, and at the endpoints the one-sided limits equal the values \sin\dfrac2\pi and \sin\dfrac{2}{3\pi}.
\text{continuous on }\left[\tfrac\pi2,\tfrac{3\pi}2\right]
15
Find the values of m and n so that the given function is continuous: f(x)=\begin{cases}mx, & x<4\\ x^{2}, & x\ge4\end{cases}.
Solution
  1. Each piece is continuous; test the joining point x=4. The value comes from x^{2} (because of \ge).
    f(4)=16,\qquad\lim_{x\to4^+}x^{2}=16,\qquad\lim_{x\to4^-}mx=4m
  2. Continuity at 4 needs the left limit equal to 16.
    4m=16\ \Longrightarrow\ m=4
  3. There is no n in this function, so only m is found.
With m = 4 the two pieces meet at (4, 16).
m=4
16
Find the values of m and n so that the given function is continuous: f(x)=\begin{cases}\dfrac{x^{2}-4}{x-2}, & x\neq2\\ m, & x=2\end{cases}.
Solution
  1. For x\neq2, f(x)=x+2, which is continuous. Find the limit at 2.
    \lim_{x\to2}\dfrac{(x-2)(x+2)}{x-2}=\lim_{x\to2}(x+2)=4
  2. Continuity at 2 needs f(2)=m to equal the limit. (There is no n in this function.)
    m=4
Defining f(2) = 4 fills the hole of the line y = x + 2.
m=4
17
Find the values of m and n so that the given function is continuous: f(x)=\begin{cases}mx, & x<3\\ n, & x=3\\ -2x+9, & x>3\end{cases}.
Solution
  1. The right-hand limit does not involve m or n.
    \lim_{x\to3^+}(-2x+9)=-6+9=3
  2. The left-hand limit must equal it.
    \lim_{x\to3^-}mx=3m=3\ \Longrightarrow\ m=1
  3. The value must equal the limit.
    f(3)=n=3
With m = 1 and n = 3 the graph is unbroken at x = 3.
m=1,\quad n=3
18
Find the values of m and n so that the given function is continuous: f(x)=\begin{cases}mx-n, & x<1\\ 5, & x=1\\ 2mx+n, & x>1\end{cases}.
Solution
  1. The value at 1 is 5, so both one-sided limits must equal 5.
    \lim_{x\to1^-}(mx-n)=m-n=5\quad(1)
  2. \lim_{x\to1^+}(2mx+n)=2m+n=5\quad(2)
  3. Add (1) and (2) to eliminate n.
    3m=10\ \Longrightarrow\ m=\dfrac{10}{3}
  4. Substitute in (1).
    n=m-5=\dfrac{10}{3}-5=-\dfrac53
  5. Check (2): 2\left(\tfrac{10}{3}\right)-\tfrac53=\tfrac{20-5}{3}=5.
Both pieces pass through (1, 5).
m=\dfrac{10}{3},\quad n=-\dfrac53
19
Prove that the equation \dfrac{x^{2}+1}{x+3}+\dfrac{x^{4}+1}{x-4}=0 has a solution in the interval (-3,4).
Solution
  1. Let f(x)=\dfrac{x^{2}+1}{x+3}+\dfrac{x^{4}+1}{x-4}. The denominators vanish only at x=-3 and x=4, so f is continuous on (-3,4), and in particular on the closed interval [0,1].
  2. Evaluate f at two convenient numbers inside (-3,4).
    f(0)=\dfrac13+\dfrac{1}{-4}=\dfrac{1}{12}>0,\qquad f(1)=\dfrac24+\dfrac{2}{-3}=\dfrac12-\dfrac23=-\dfrac16<0
  3. f is continuous on [0,1] and changes sign, so by the intermediate value theorem there is a number c in (0,1) with f(c)=0.
  4. Since (0,1)\subset(-3,4), the equation has a solution in (-3,4). (Numerically c\approx0.78.)
f\to+\infty as x\to-3^{+} and f\to-\infty as x\to4^{-}; the continuous graph must cross the x-axis in between.
f(0)>0,\ f(1)<0\ \Longrightarrow\ \exists\,c\in(0,1)\subset(-3,4):\ f(c)=0
20
Prove that f(x)=\begin{cases}1, & x\ \text{rational}\\ 0, & x\ \text{irrational}\end{cases} is discontinuous at every real number. What does the graph of f look like?
Solution
  1. Take any real number a. Every interval around a, however small, contains both rational and irrational numbers (they are dense in the real line).
  2. So in every neighbourhood of a the function takes the value 1 (at rational points) and the value 0 (at irrational points). As x\to a, f(x) keeps jumping between 0 and 1 and cannot get close to any single number L: it cannot be within \tfrac12 of both 0 and 1.
  3. Hence \lim_{x\to a}f(x) does not exist, condition (ii) fails, and f is discontinuous at a. Since a was any real number, f is discontinuous everywhere.
    \lim_{x\to a}f(x)\ \text{does not exist for every }a\in\mathbb{R}
  4. Graph. Above each rational x there is a point at height 1, and above each irrational x a point at height 0. The graph looks like two dotted horizontal lines, y=1 and y=0, each full of gaps that are too small to see; it cannot be drawn properly.
A sketch only: both dotted lines are really infinitely many separate points.
f\ \text{is discontinuous at every real number}

Exercise 2.3p. 55

Slopes of tangents and instantaneous velocities are found from the definition with the four steps of Section 2.3: find f(a) and f(a+\Delta x), then \Delta y, then \dfrac{\Delta y}{\Delta x}, and finally \displaystyle\lim_{\Delta x\to0}\dfrac{\Delta y}{\Delta x}. An average rate of change is the slope of a secant: \dfrac{f(b)-f(a)}{b-a}.

1
Find the slope of the tangent line to the graph of the given function at the indicated point: f(x)=2x-1; (x,f(x))=(4,7).
Solution
  1. Step 1. Evaluate f at a=4 and at 4+\Delta x, where \Delta x\neq0.
    f(4)=2(4)-1=7\qquad f(4+\Delta x)=2(4+\Delta x)-1=7+2\Delta x
  2. Step 2. Subtract to find the change in y: \Delta y=f(a+\Delta x)-f(a).
    \Delta y=f(4+\Delta x)-f(4)=(7+2\Delta x)-7=2\Delta x
  3. Step 3. Divide by \Delta x. This is allowed because \Delta x\neq0.
    \dfrac{\Delta y}{\Delta x}=\dfrac{2\Delta x}{\Delta x}=2
  4. Step 4. Let \Delta x\to0. The quotient is the constant 2, so its limit is 2.
    m_{\tan}=\lim_{\Delta x\to0}\dfrac{\Delta y}{\Delta x}=\lim_{\Delta x\to0}2=2
m_{\tan}=2
2
Find the slope of the tangent line to the graph of the given function at the indicated point: f(x)=-\dfrac{1}{2}x+3; (a,f(a)).
Solution
  1. Step 1. Here the point is general, x=a. Evaluate f at a and at a+\Delta x.
    f(a)=-\dfrac{1}{2}a+3\qquad f(a+\Delta x)=-\dfrac{1}{2}(a+\Delta x)+3=-\dfrac{1}{2}a-\dfrac{1}{2}\Delta x+3
  2. Step 2. Find \Delta y. The terms -\tfrac12a and 3 cancel.
    \Delta y=\left(-\dfrac{1}{2}a-\dfrac{1}{2}\Delta x+3\right)-\left(-\dfrac{1}{2}a+3\right)=-\dfrac{1}{2}\Delta x
  3. Step 3. Divide by \Delta x. This is allowed because \Delta x\neq0.
    \dfrac{\Delta y}{\Delta x}=\dfrac{-\frac{1}{2}\Delta x}{\Delta x}=-\dfrac{1}{2}
  4. Step 4. Let \Delta x\to0. The limit is the slope of the tangent line.
    m_{\tan}=\lim_{\Delta x\to0}\left(-\dfrac{1}{2}\right)=-\dfrac{1}{2}
m_{\tan}=-\dfrac{1}{2}
3
Find the slope of the tangent line to the graph of the given function at the indicated point: f(x)=x^{2}+4; (-1,5).
Solution
  1. Step 1. Evaluate f at a=-1 and at -1+\Delta x. Expand the square with (p+q)^{2}=p^{2}+2pq+q^{2}.
    \begin{aligned}f(-1)&=(-1)^{2}+4=5\\ f(-1+\Delta x)&=(-1+\Delta x)^{2}+4\\ &=1-2\Delta x+(\Delta x)^{2}+4=5-2\Delta x+(\Delta x)^{2}\end{aligned}
  2. Step 2. Subtract f(-1)=5 and take out the common factor \Delta x.
    \Delta y=\left[5-2\Delta x+(\Delta x)^{2}\right]-5=-2\Delta x+(\Delta x)^{2}=\Delta x(-2+\Delta x)
  3. Step 3. Divide by \Delta x. This is allowed because \Delta x\neq0.
    \dfrac{\Delta y}{\Delta x}=\dfrac{\Delta x(-2+\Delta x)}{\Delta x}=-2+\Delta x
  4. Step 4. Let \Delta x\to0. The limit is the slope of the tangent line.
    m_{\tan}=\lim_{\Delta x\to0}(-2+\Delta x)=-2
  5. The tangent line passes through (-1,5) with slope -2. Its equation is shown on the graph.
    y-5=-2(x+1)\;\Longrightarrow\;y=-2x+3
The tangent at (-1,5) touches the parabola and slopes downwards, matching m_{\tan}=-2.
m_{\tan}=-2
4
Find the slope of the tangent line to the graph of the given function at the indicated point: f(x)=x^{2}-5x+4; (2,-2).
Solution
  1. Step 1. Evaluate f at a=2 (this confirms the point is on the graph) and at 2+\Delta x.
    \begin{aligned}f(2)&=2^{2}-5(2)+4=-2\\ f(2+\Delta x)&=(2+\Delta x)^{2}-5(2+\Delta x)+4\\ &=4+4\Delta x+(\Delta x)^{2}-10-5\Delta x+4\\ &=-2-\Delta x+(\Delta x)^{2}\end{aligned}
  2. Step 2. Subtract f(2)=-2 and factorise.
    \Delta y=\left[-2-\Delta x+(\Delta x)^{2}\right]-(-2)=-\Delta x+(\Delta x)^{2}=\Delta x(-1+\Delta x)
  3. Step 3. Divide by \Delta x. This is allowed because \Delta x\neq0.
    \dfrac{\Delta y}{\Delta x}=\dfrac{\Delta x(-1+\Delta x)}{\Delta x}=-1+\Delta x
  4. Step 4. Let \Delta x\to0. The limit is the slope of the tangent line.
    m_{\tan}=\lim_{\Delta x\to0}(-1+\Delta x)=-1
  5. The tangent line through (2,-2) with slope -1:
    y+2=-1(x-2)\;\Longrightarrow\;y=-x
The point (2,-2) is just left of the vertex (2.5,-2.25), so the tangent there slopes gently downwards.
m_{\tan}=-1
5
Find the slope of the tangent line to the graph of the given function at the indicated point: f(x)=x^{3}; (1,f(1)).
Solution
  1. Step 1. Evaluate f at a=1 and at 1+\Delta x. Expand the cube with (p+q)^{3}=p^{3}+3p^{2}q+3pq^{2}+q^{3}.
    \begin{aligned}f(1)&=1^{3}=1\\ f(1+\Delta x)&=(1+\Delta x)^{3}=1+3\Delta x+3(\Delta x)^{2}+(\Delta x)^{3}\end{aligned}
  2. Step 2. Subtract f(1)=1 and take out the factor \Delta x.
    \Delta y=\left[1+3\Delta x+3(\Delta x)^{2}+(\Delta x)^{3}\right]-1=\Delta x\left[3+3\Delta x+(\Delta x)^{2}\right]
  3. Step 3. Divide by \Delta x. This is allowed because \Delta x\neq0.
    \dfrac{\Delta y}{\Delta x}=3+3\Delta x+(\Delta x)^{2}
  4. Step 4. Let \Delta x\to0: both terms containing \Delta x vanish.
    m_{\tan}=\lim_{\Delta x\to0}\left[3+3\Delta x+(\Delta x)^{2}\right]=3
  5. The point is (1,1), so the tangent line is:
    y-1=3(x-1)\;\Longrightarrow\;y=3x-2
The tangent at (1,1) cuts the curve again at (-2,-8). A tangent only has to touch the curve at the point of contact.
m_{\tan}=3
6
Find the slope of the tangent line to the graph of the given function at the indicated point: f(x)=\dfrac{1}{x}; \left(\dfrac{1}{3},f\left(\dfrac{1}{3}\right)\right).
Solution
  1. Step 1. Evaluate f at a=\tfrac13 and at \tfrac13+\Delta x. Multiply the top and bottom by 3 to clear the small fraction.
    f\left(\tfrac{1}{3}\right)=\dfrac{1}{1/3}=3\qquad f\left(\tfrac{1}{3}+\Delta x\right)=\dfrac{1}{\frac{1}{3}+\Delta x}=\dfrac{3}{1+3\Delta x}
  2. Step 2. Subtract f(\tfrac13)=3 using the common denominator 1+3\Delta x.
    \Delta y=\dfrac{3}{1+3\Delta x}-3=\dfrac{3-3(1+3\Delta x)}{1+3\Delta x}=\dfrac{-9\Delta x}{1+3\Delta x}
  3. Step 3. Divide by \Delta x. This is allowed because \Delta x\neq0.
    \dfrac{\Delta y}{\Delta x}=\dfrac{-9}{1+3\Delta x}
  4. Step 4. Let \Delta x\to0; the denominator tends to 1.
    m_{\tan}=\lim_{\Delta x\to0}\dfrac{-9}{1+3\Delta x}=\dfrac{-9}{1+0}=-9
  5. The point is \left(\tfrac13,3\right), so the tangent line is:
    y-3=-9\left(x-\tfrac{1}{3}\right)\;\Longrightarrow\;y=-9x+6
Near x=\tfrac13 the curve y=\dfrac1x falls very steeply, so the slope -9 is large and negative.
m_{\tan}=-9
7
Find the average rate of change of the given function in the indicated interval: f(x)=x^{3}+2x^{2}-4x; [-1,2].
Solution
  1. The average rate of change of f on [a,b] is the slope of the secant line joining (a,f(a)) and (b,f(b)). Here a=-1 and b=2.
    \dfrac{\Delta y}{\Delta x}=\dfrac{f(b)-f(a)}{b-a}=\dfrac{f(2)-f(-1)}{2-(-1)}
  2. Evaluate f at the left end point x=-1.
    f(-1)=(-1)^{3}+2(-1)^{2}-4(-1)=-1+2+4=5
  3. Evaluate f at the right end point x=2.
    f(2)=2^{3}+2(2)^{2}-4(2)=8+8-8=8
  4. Divide the change in f by the length of the interval.
    \dfrac{\Delta y}{\Delta x}=\dfrac{8-5}{2-(-1)}=\dfrac{3}{3}=1
The secant through (-1,5) and (2,8) has slope 1. The curve dips below it in between, but on average f rises 1 unit for each unit of x.
\dfrac{\Delta y}{\Delta x}=1
8
Find the average rate of change of the given function in the indicated interval: f(x)=\cos x; [-\pi,\pi].
Solution
  1. Use the same formula with a=-\pi and b=\pi.
    \dfrac{\Delta y}{\Delta x}=\dfrac{f(\pi)-f(-\pi)}{\pi-(-\pi)}
  2. Evaluate f at both end points. Cosine is an even function, \cos(-\theta)=\cos\theta, so the two values are equal.
    f(-\pi)=\cos(-\pi)=\cos\pi=-1\qquad f(\pi)=\cos\pi=-1
  3. Divide the change in f by the length 2\pi of the interval.
    \dfrac{\Delta y}{\Delta x}=\dfrac{-1-(-1)}{2\pi}=\dfrac{0}{2\pi}=0
The secant joining the end points is horizontal, so the average rate of change is 0.
\dfrac{\Delta y}{\Delta x}=0
9
Find the instantaneous velocity of the particle at the indicated time: f(t)=-4t^{2}+10t+6; t=3.
Solution

The instantaneous velocity is v(t_{1})=\displaystyle\lim_{\Delta t\to0}\dfrac{f(t_{1}+\Delta t)-f(t_{1})}{\Delta t}. Use the same four steps with t_{1}=3.

  1. Step 1. Evaluate the position at t=3 and at 3+\Delta t, where \Delta t\neq0.
    \begin{aligned}f(3)&=-4(3)^{2}+10(3)+6=-36+30+6=0\\ f(3+\Delta t)&=-4(3+\Delta t)^{2}+10(3+\Delta t)+6\\ &=-36-24\Delta t-4(\Delta t)^{2}+30+10\Delta t+6\\ &=-14\Delta t-4(\Delta t)^{2}\end{aligned}
  2. Step 2. Find the change in position \Delta s.
    \Delta s=f(3+\Delta t)-f(3)=-14\Delta t-4(\Delta t)^{2}-0=\Delta t(-14-4\Delta t)
  3. Step 3. Divide by \Delta t\neq0.
    \dfrac{\Delta s}{\Delta t}=-14-4\Delta t
  4. Step 4. Let \Delta t\to0.
    v(3)=\lim_{\Delta t\to0}(-14-4\Delta t)=-14
v(3)=-14
10
Find the instantaneous velocity of the particle at the indicated time: f(t)=t^{2}+\dfrac{1}{5t+1}; t=0.
Solution
  1. Step 1. Evaluate the position at t=0 and at 0+\Delta t=\Delta t.
    f(0)=0^{2}+\dfrac{1}{5(0)+1}=1\qquad f(\Delta t)=(\Delta t)^{2}+\dfrac{1}{1+5\Delta t}
  2. Step 2. Subtract f(0)=1, writing \dfrac{1}{1+5\Delta t}-1 over one denominator.
    \begin{aligned}\Delta s&=(\Delta t)^{2}+\dfrac{1}{1+5\Delta t}-1\\ &=(\Delta t)^{2}+\dfrac{1-(1+5\Delta t)}{1+5\Delta t}=(\Delta t)^{2}-\dfrac{5\Delta t}{1+5\Delta t}\end{aligned}
  3. Step 3. Divide each term by \Delta t\neq0.
    \dfrac{\Delta s}{\Delta t}=\Delta t-\dfrac{5}{1+5\Delta t}
  4. Step 4. Let \Delta t\to0.
    v(0)=\lim_{\Delta t\to0}\left(\Delta t-\dfrac{5}{1+5\Delta t}\right)=0-\dfrac{5}{1+0}=-5
v(0)=-5
11
The height above ground of a ball dropped from an initial altitude of 122.5\,\mathrm{m} is given by s(t)=122.5-4.9t^{2}, where s is measured in meters and t in seconds.
Solution
(i) What is the instantaneous velocity at t=\dfrac{1}{2}?
  1. Find the velocity at any time t first (this is reused in part (iii)). Step 1: write the height at time t+\Delta t and expand.
    \begin{aligned}s(t+\Delta t)&=122.5-4.9(t+\Delta t)^{2}\\ &=122.5-4.9t^{2}-9.8t\,\Delta t-4.9(\Delta t)^{2}\end{aligned}
  2. Step 2: subtract s(t)=122.5-4.9t^{2}.
    \begin{aligned}\Delta s&=s(t+\Delta t)-s(t)=-9.8t\,\Delta t-4.9(\Delta t)^{2}\\ &=\Delta t\,(-9.8t-4.9\Delta t)\end{aligned}
  3. Step 3: divide by \Delta t\neq0.
    \dfrac{\Delta s}{\Delta t}=-9.8t-4.9\Delta t
  4. Step 4: let \Delta t\to0.
    v(t)=\lim_{\Delta t\to0}(-9.8t-4.9\Delta t)=-9.8t
  5. Put t=\tfrac12.
    v\left(\tfrac{1}{2}\right)=-9.8\left(\tfrac{1}{2}\right)=-4.9\,\mathrm{m\,s^{-1}}
v\left(\tfrac{1}{2}\right)=-4.9\,\mathrm{m\,s^{-1}}
(ii) At what time does the ball hit the ground?
  1. The ball is on the ground when its height is zero.
    s(t)=0\;\Longrightarrow\;122.5-4.9t^{2}=0
  2. Make t^{2} the subject.
    4.9t^{2}=122.5\;\Longrightarrow\;t^{2}=\dfrac{122.5}{4.9}=25
  3. Take the square root. Time after the drop cannot be negative, so reject t=-5.
    t=\sqrt{25}=5\,\mathrm{s}
t=5\,\mathrm{s}
(iii) What is the impact velocity?
  1. The impact velocity is the instantaneous velocity at the moment the ball reaches the ground, t=5\,\mathrm{s} from part (ii). Use v(t)=-9.8t from part (i).
    v(5)=-9.8(5)=-49\,\mathrm{m\,s^{-1}}

The ball hits the ground moving downwards with a speed of 49\,\mathrm{m\,s^{-1}}.

Height against time. The curve gets steeper as the ball falls, so its speed increases until impact at t=5\,\mathrm{s}.
v(5)=-49\,\mathrm{m\,s^{-1}}

Note. Part (i) is printed as "at =\tfrac12"; the missing symbol is t, so it asks for the velocity at t=\tfrac12\,\mathrm{s}.

12
The height of a projectile shot from ground level is given by s(t)=-16t^{2}+256t, where s is measured in feet and t in seconds.
Solution
(i) Determine the height of the projectile at t=2, t=6, t=9 and t=10.
  1. Substitute each time into s(t)=-16t^{2}+256t. At t=2:
    s(2)=-16(4)+256(2)=-64+512=448\,\mathrm{ft}
  2. At t=6:
    s(6)=-16(36)+256(6)=-576+1536=960\,\mathrm{ft}
  3. At t=9:
    s(9)=-16(81)+256(9)=-1296+2304=1008\,\mathrm{ft}
  4. At t=10 the height is again 960\,\mathrm{ft}, the same as at t=6: the projectile is now on its way down.
    s(10)=-16(100)+256(10)=-1600+2560=960\,\mathrm{ft}
\begin{aligned}s(2)&=448\,\mathrm{ft}\\ s(6)&=960\,\mathrm{ft}\\ s(9)&=1008\,\mathrm{ft}\\ s(10)&=960\,\mathrm{ft}\end{aligned}
(ii) What is the average velocity of the projectile between t=2 and t=5?
  1. Average velocity is the change in position divided by the time taken.
    v_{\text{ave}}=\dfrac{s(5)-s(2)}{5-2}
  2. Find the height at t=5. The height at t=2 is 448\,\mathrm{ft} from part (i).
    s(5)=-16(25)+256(5)=-400+1280=880\,\mathrm{ft}
  3. Substitute and divide.
    v_{\text{ave}}=\dfrac{880-448}{3}=\dfrac{432}{3}=144\,\mathrm{ft\,s^{-1}}
v_{\text{ave}}=144\,\mathrm{ft\,s^{-1}}
(iii) Show that the average velocity between t=7 and t=9 is zero, also interpret.
  1. Find the heights at the two times (s(9) is from part (i)).
    \begin{aligned}s(7)&=-16(49)+256(7)=-784+1792=1008\,\mathrm{ft}\\ s(9)&=1008\,\mathrm{ft}\end{aligned}
  2. Divide the change in height by the time interval.
    v_{\text{ave}}=\dfrac{s(9)-s(7)}{9-7}=\dfrac{1008-1008}{2}=0
  3. Interpretation. The projectile is at the same height, 1008\,\mathrm{ft}, at t=7 and at t=9. Between these times it rises to its highest point (at t=8) and then falls back by the same amount. Its net change in position is zero, so its average velocity is zero, even though it was moving all the time.
The secant joining the points at t=7 and t=9 is horizontal, so the average velocity over this interval is zero.
v_{\text{ave}}=0
(iv) At what time does the projectile hit the ground?
  1. The projectile is on the ground when its height is zero.
    -16t^{2}+256t=0
  2. Take out the common factor -16t.
    -16t(t-16)=0\;\Longrightarrow\;t=0\ \text{or}\ t=16
  3. t=0 is the moment of launch from ground level, so the projectile lands at the other root.
    t=16\,\mathrm{s}
t=16\,\mathrm{s}
(v) Determine the instantaneous velocity at time t=8.
  1. Step 1. Evaluate s at t=8 and at 8+\Delta t. The terms \pm256\Delta t cancel.
    \begin{aligned}s(8)&=-16(64)+256(8)=-1024+2048=1024\\ s(8+\Delta t)&=-16(8+\Delta t)^{2}+256(8+\Delta t)\\ &=-1024-256\Delta t-16(\Delta t)^{2}+2048+256\Delta t\\ &=1024-16(\Delta t)^{2}\end{aligned}
  2. Step 2. Change in position.
    \Delta s=s(8+\Delta t)-s(8)=1024-16(\Delta t)^{2}-1024=-16(\Delta t)^{2}
  3. Step 3. Divide by \Delta t\neq0.
    \dfrac{\Delta s}{\Delta t}=\dfrac{-16(\Delta t)^{2}}{\Delta t}=-16\Delta t
  4. Step 4. Let \Delta t\to0.
    v(8)=\lim_{\Delta t\to0}(-16\Delta t)=0\,\mathrm{ft\,s^{-1}}
v(8)=0\,\mathrm{ft\,s^{-1}}
(vi) What is the maximum height that the projectile attains?
  1. At the highest point the projectile stops going up before it starts coming down, so its instantaneous velocity there is zero. Find v(t) at any time with the four-step method: first the change in position.
    \Delta s=s(t+\Delta t)-s(t)=-32t\,\Delta t-16(\Delta t)^{2}+256\Delta t
  2. Divide by \Delta t and let \Delta t\to0.
    v(t)=\lim_{\Delta t\to0}(-32t+256-16\Delta t)=-32t+256
  3. Set the velocity equal to zero (this agrees with part (v)).
    -32t+256=0\;\Longrightarrow\;t=8\,\mathrm{s}
  4. Find the height at this time.
    s(8)=-16(8)^{2}+256(8)=-1024+2048=1024\,\mathrm{ft}
  5. Check by completing the square: the height can never be more than 1024, and it equals 1024 only at t=8.
    s(t)=-16\left(t^{2}-16t\right)=-16\left[(t-8)^{2}-64\right]=1024-16(t-8)^{2}
The height rises to a maximum of 1024\,\mathrm{ft} at t=8\,\mathrm{s} and the projectile lands at t=16\,\mathrm{s}.
s_{\max}=1024\,\mathrm{ft}

Exercise 2.4p. 59–60

Use the rules of Section 2.6: \dfrac{d}{dx}[c]=0, \dfrac{d}{dx}x^{n}=nx^{n-1} for any real n, \dfrac{d}{dx}[cf(x)]=cf'(x) and \dfrac{d}{dx}[f(x)\pm g(x)]=f'(x)\pm g'(x). Write roots as fractional powers and multiply out products before differentiating.

1
Find the derivative of the functions w.r.t. x.
Solution
(a) y=x^{9}
  1. Use the power rule \dfrac{d}{dx}x^{n}=nx^{n-1}: bring the power down as a factor and reduce the power by 1. Here n=9.
    \dfrac{dy}{dx}=\dfrac{d}{dx}x^{9}=9x^{9-1}=9x^{8}
\dfrac{dy}{dx}=9x^{8}
(b) f(x)=4x^{\frac{1}{3}}
  1. The constant 4 stays as a multiplier (constant multiple rule, \dfrac{d}{dx}[cf(x)]=cf'(x)).
    f'(x)=4\,\dfrac{d}{dx}x^{\frac{1}{3}}
  2. Apply the power rule with n=\tfrac13. The new power is \tfrac13-1=-\tfrac23.
    f'(x)=4\cdot\dfrac{1}{3}x^{\frac{1}{3}-1}=\dfrac{4}{3}x^{-\frac{2}{3}}
  3. Write the answer with a positive power, using x^{\frac23}=\sqrt[3]{x^{2}}.
    f'(x)=\dfrac{4}{3x^{\frac{2}{3}}}=\dfrac{4}{3\sqrt[3]{x^{2}}}
f'(x)=\dfrac{4}{3}x^{-\frac{2}{3}}=\dfrac{4}{3\sqrt[3]{x^{2}}}
(c) f(x)=9
  1. The derivative of a constant is zero: a constant does not change when x changes. The graph is the horizontal line y=9, whose slope is 0 everywhere.
    f'(x)=\dfrac{d}{dx}(9)=0
f'(x)=0
(d) f(x)=6x^{3}+3x^{2}-10
  1. Differentiate term by term (sum and difference rule), keeping each constant multiplier.
    f'(x)=6\dfrac{d}{dx}x^{3}+3\dfrac{d}{dx}x^{2}-\dfrac{d}{dx}(10)
  2. Apply the power rule to each term; the constant term 10 gives 0.
    f'(x)=6(3x^{2})+3(2x)-0=18x^{2}+6x
f'(x)=18x^{2}+6x
2
Determine f'(x).
Solution
(a) f(x)=\sqrt{5}
  1. \sqrt5\approx2.236 is a fixed number, so f is a constant function and its derivative is zero.
    f'(x)=\dfrac{d}{dx}\sqrt{5}=0
f'(x)=0
(b) f(x)=\sqrt{5}\,x
  1. \sqrt5 is a constant multiplier and \dfrac{d}{dx}x=1.
    f'(x)=\sqrt{5}\,\dfrac{d}{dx}x=\sqrt{5}(1)=\sqrt{5}
f'(x)=\sqrt{5}
(c) f(x)=5\sqrt{x}
  1. Write the root as a power: \sqrt{x}=x^{\frac12}.
    f(x)=5x^{\frac{1}{2}}
  2. Apply the power rule with n=\tfrac12, keeping the multiplier 5.
    f'(x)=5\cdot\dfrac{1}{2}x^{\frac{1}{2}-1}=\dfrac{5}{2}x^{-\frac{1}{2}}=\dfrac{5}{2\sqrt{x}}
f'(x)=\dfrac{5}{2\sqrt{x}}
(d) f(x)=\sqrt{5x}
  1. Separate the constant with \sqrt{ab}=\sqrt{a}\,\sqrt{b}.
    f(x)=\sqrt{5}\,\sqrt{x}=\sqrt{5}\,x^{\frac{1}{2}}
  2. Apply the power rule with n=\tfrac12; \sqrt5 is a constant multiplier.
    f'(x)=\sqrt{5}\cdot\dfrac{1}{2}x^{-\frac{1}{2}}=\dfrac{\sqrt{5}}{2\sqrt{x}}
f'(x)=\dfrac{\sqrt{5}}{2\sqrt{x}}
3
Determine f'(x).
Solution
(a) f(x)=x^{2}(x^{3}+5)
  1. Multiply out first, so that only the power rule is needed.
    f(x)=x^{5}+5x^{2}
  2. Differentiate term by term.
    f'(x)=5x^{4}+5(2x)=5x^{4}+10x
f'(x)=5x^{4}+10x
(b) f(x)=(x+9)(x-9)
  1. Use the identity (a+b)(a-b)=a^{2}-b^{2}.
    f(x)=x^{2}-81
  2. The constant 81 has derivative 0.
    f'(x)=2x-0=2x
f'(x)=2x
(c) f(x)=(x^{2}+x^{3})^{3}
  1. Take out the common factor x^{2} inside the bracket, then cube each factor.
    f(x)=\left[x^{2}(1+x)\right]^{3}=x^{6}(1+x)^{3}
  2. Expand (1+x)^{3}=1+3x+3x^{2}+x^{3} and multiply by x^{6}.
    f(x)=x^{6}+3x^{7}+3x^{8}+x^{9}
  3. Apply the power rule to each term.
    f'(x)=6x^{5}+21x^{6}+24x^{7}+9x^{8}
  4. The answer factorises neatly; this is the form the power rule for functions gives directly.
    f'(x)=3x^{5}(1+x)^{2}(2+3x)
f'(x)=6x^{5}+21x^{6}+24x^{7}+9x^{8}=3x^{5}(1+x)^{2}(2+3x)
(d) f(x)=-3x^{-8}+2\sqrt{x}
  1. Write the root as a power.
    f(x)=-3x^{-8}+2x^{\frac{1}{2}}
  2. Apply the power rule to each term: the new powers are -8-1=-9 and \tfrac12-1=-\tfrac12.
    f'(x)=-3(-8)x^{-9}+2\cdot\dfrac{1}{2}x^{-\frac{1}{2}}=24x^{-9}+x^{-\frac{1}{2}}
  3. Write the answer with positive powers.
    f'(x)=\dfrac{24}{x^{9}}+\dfrac{1}{\sqrt{x}}
f'(x)=\dfrac{24}{x^{9}}+\dfrac{1}{\sqrt{x}}
(e) f(x)=ax^{3}+bx^{2}+cx+d, (a, b, c and d are constants)
  1. Treat a, b, c like numbers: they stay as multipliers. The constant term d has derivative 0.
    f'(x)=a(3x^{2})+b(2x)+c(1)+0
  2. Simplify.
    f'(x)=3ax^{2}+2bx+c
f'(x)=3ax^{2}+2bx+c
(f) f(x)=x^{24}+2x^{\frac{1}{2}}+3x^{8}+9x^{4}
  1. Apply the power rule to each term, keeping the multipliers.
    f'(x)=24x^{23}+2\cdot\dfrac{1}{2}x^{-\frac{1}{2}}+3(8x^{7})+9(4x^{3})
  2. Simplify, writing x^{-\frac12}=\dfrac{1}{\sqrt{x}}.
    f'(x)=24x^{23}+\dfrac{1}{\sqrt{x}}+24x^{7}+36x^{3}
f'(x)=24x^{23}+\dfrac{1}{\sqrt{x}}+24x^{7}+36x^{3}
4
Find \dfrac{dy}{dx}.
Solution
(a) y=\dfrac{x+2x^{\frac{3}{2}}}{\sqrt{x}}
  1. Divide each term of the numerator by \sqrt{x}=x^{\frac12}, using x^{m}\div x^{n}=x^{m-n}.
    y=\dfrac{x}{x^{\frac{1}{2}}}+\dfrac{2x^{\frac{3}{2}}}{x^{\frac{1}{2}}}=x^{\frac{1}{2}}+2x
  2. Differentiate term by term.
    \dfrac{dy}{dx}=\dfrac{1}{2}x^{-\frac{1}{2}}+2=\dfrac{1}{2\sqrt{x}}+2
\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}+2
(b) y=(x^{3}-5)(2x+3)
  1. Multiply out the brackets.
    y=2x^{4}+3x^{3}-10x-15
  2. Differentiate term by term; the constant -15 gives 0.
    \dfrac{dy}{dx}=8x^{3}+9x^{2}-10
\dfrac{dy}{dx}=8x^{3}+9x^{2}-10
(c) y=(4x^{2}-3)(7x^{2}+x)
  1. Multiply out the brackets.
    y=28x^{4}+4x^{3}-21x^{2}-3x
  2. Differentiate term by term.
    \dfrac{dy}{dx}=112x^{3}+12x^{2}-42x-3
\dfrac{dy}{dx}=112x^{3}+12x^{2}-42x-3
5
Find slope of tangent at x=1.
Solution
(a) y=x^{2}+3x
  1. The slope of the tangent at a point is the value of the derivative there. First differentiate.
    \dfrac{dy}{dx}=2x+3
  2. Substitute x=1.
    m=\left.\dfrac{dy}{dx}\right|_{x=1}=2(1)+3=5
  3. The point of contact is (1,\,1+3)=(1,4), so the tangent line is y-4=5(x-1), that is y=5x-1.
The tangent at (1,4) rises steeply with slope 5.
m=5
(b) y=x^{4}-x^{2}
  1. Differentiate with the power rule.
    \dfrac{dy}{dx}=4x^{3}-2x
  2. Substitute x=1.
    m=4(1)^{3}-2(1)=2
  3. The point of contact is (1,\,1-1)=(1,0), so the tangent line is y=2(x-1), that is y=2x-2.
At (1,0) the curve is rising, and the tangent has slope 2.
m=2

Exercise 2.5p. 62

Use the product rule \dfrac{d}{dx}[f\cdot g]=fg'+gf' and the quotient rule \dfrac{d}{dx}\left[\dfrac{f}{g}\right]=\dfrac{gf'-fg'}{g^{2}} (Section 2.7). For each question, first write f, f', g and g', then substitute and simplify the numerator.

1
Find \dfrac{dy}{dx} if y=\dfrac{1}{x}.
Solution
  1. Use the quotient rule with numerator f=1 and denominator g=x.
    \begin{aligned}f&=1\\ f'&=0\\ g&=x\\ g'&=1\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}: denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared.
    \dfrac{dy}{dx}=\dfrac{x(0)-1(1)}{x^{2}}=-\dfrac{1}{x^{2}}
  3. Check with the power rule: y=x^{-1}.
    \dfrac{dy}{dx}=-1\cdot x^{-2}=-\dfrac{1}{x^{2}}
\dfrac{dy}{dx}=-\dfrac{1}{x^{2}}
2
Find \dfrac{dy}{dx} if y=(x^{2}-7)(x^{2}+4x+2).
Solution
  1. The function is a product. Name the two factors f and g, and differentiate each.
    \begin{aligned}f&=x^{2}-7\\ f'&=2x\\ g&=x^{2}+4x+2\\ g'&=2x+4\end{aligned}
  2. Apply the product rule \dfrac{dy}{dx}=fg'+gf': first function times derivative of the second, plus second function times derivative of the first.
    \dfrac{dy}{dx}=(x^{2}-7)(2x+4)+(x^{2}+4x+2)(2x)
  3. Expand both products.
    \dfrac{dy}{dx}=2x^{3}+4x^{2}-14x-28+2x^{3}+8x^{2}+4x
  4. Collect like terms.
    \dfrac{dy}{dx}=4x^{3}+12x^{2}-10x-28
\dfrac{dy}{dx}=4x^{3}+12x^{2}-10x-28
3
Find \dfrac{dy}{dx} if y=(7x+1)(x^{4}-x^{3}-9x).
Solution
  1. The function is a product. Name the two factors f and g, and differentiate each.
    \begin{aligned}f&=7x+1\\ f'&=7\\ g&=x^{4}-x^{3}-9x\\ g'&=4x^{3}-3x^{2}-9\end{aligned}
  2. Apply the product rule \dfrac{dy}{dx}=fg'+gf': first function times derivative of the second, plus second function times derivative of the first.
    \dfrac{dy}{dx}=(7x+1)(4x^{3}-3x^{2}-9)+(x^{4}-x^{3}-9x)(7)
  3. Expand both products.
    \dfrac{dy}{dx}=28x^{4}-21x^{3}-63x+4x^{3}-3x^{2}-9+7x^{4}-7x^{3}-63x
  4. Collect like terms.
    \dfrac{dy}{dx}=35x^{4}-24x^{3}-3x^{2}-126x-9
\dfrac{dy}{dx}=35x^{4}-24x^{3}-3x^{2}-126x-9
4
Find \dfrac{dy}{dx} if y=\dfrac{3x+4}{x^{2}+1}.
Solution
  1. The function is a quotient. Name the numerator f and the denominator g, and differentiate each.
    \begin{aligned}f&=3x+4\\ f'&=3\\ g&=x^{2}+1\\ g'&=2x\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}: denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared.
    \dfrac{dy}{dx}=\dfrac{(x^{2}+1)(3)-(3x+4)(2x)}{(x^{2}+1)^{2}}
  3. Expand the numerator and collect like terms. Leave the denominator as a square.
    \dfrac{dy}{dx}=\dfrac{3x^{2}+3-6x^{2}-8x}{(x^{2}+1)^{2}}=\dfrac{-3x^{2}-8x+3}{(x^{2}+1)^{2}}
\dfrac{dy}{dx}=\dfrac{-3x^{2}-8x+3}{(x^{2}+1)^{2}}
5
Find \dfrac{dy}{dx} if y=\dfrac{x-2}{x^{4}+x+1}.
Solution
  1. The function is a quotient. Name the numerator f and the denominator g, and differentiate each.
    \begin{aligned}f&=x-2\\ f'&=1\\ g&=x^{4}+x+1\\ g'&=4x^{3}+1\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}: denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared.
    \dfrac{dy}{dx}=\dfrac{(x^{4}+x+1)(1)-(x-2)(4x^{3}+1)}{(x^{4}+x+1)^{2}}
  3. Expand (x-2)(4x^{3}+1)=4x^{4}-8x^{3}+x-2 and subtract it; the x terms cancel.
    \begin{aligned}\dfrac{dy}{dx}&=\dfrac{x^{4}+x+1-4x^{4}+8x^{3}-x+2}{(x^{4}+x+1)^{2}}\\ &=\dfrac{-3x^{4}+8x^{3}+3}{(x^{4}+x+1)^{2}}\end{aligned}
\dfrac{dy}{dx}=\dfrac{-3x^{4}+8x^{3}+3}{(x^{4}+x+1)^{2}}
6
Find \dfrac{dy}{dx} if y=\dfrac{3x^{2}+5}{3x-1}.
Solution
  1. The function is a quotient. Name the numerator f and the denominator g, and differentiate each.
    \begin{aligned}f&=3x^{2}+5\\ f'&=6x\\ g&=3x-1\\ g'&=3\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}: denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared.
    \dfrac{dy}{dx}=\dfrac{(3x-1)(6x)-(3x^{2}+5)(3)}{(3x-1)^{2}}
  3. Expand the numerator and collect like terms.
    \dfrac{dy}{dx}=\dfrac{18x^{2}-6x-9x^{2}-15}{(3x-1)^{2}}=\dfrac{9x^{2}-6x-15}{(3x-1)^{2}}
  4. Take out the common factor 3 and factorise the quadratic.
    \dfrac{dy}{dx}=\dfrac{3(3x^{2}-2x-5)}{(3x-1)^{2}}=\dfrac{3(3x-5)(x+1)}{(3x-1)^{2}}
\dfrac{dy}{dx}=\dfrac{9x^{2}-6x-15}{(3x-1)^{2}}=\dfrac{3(3x-5)(x+1)}{(3x-1)^{2}}
7
Find \dfrac{dy}{dx} if y=\left(\dfrac{1}{x}+\dfrac{1}{x^{2}}\right)(3x^{3}+27).
Solution
  1. The function is a product. Write the first factor with negative powers, which are easy to differentiate.
    \begin{aligned}f&=x^{-1}+x^{-2}\\ f'&=-x^{-2}-2x^{-3}\\ g&=3x^{3}+27\\ g'&=9x^{2}\end{aligned}
  2. Apply the product rule \dfrac{dy}{dx}=fg'+gf': first function times derivative of the second, plus second function times derivative of the first.
    \dfrac{dy}{dx}=\left(\dfrac{1}{x}+\dfrac{1}{x^{2}}\right)(9x^{2})+(3x^{3}+27)\left(-\dfrac{1}{x^{2}}-\dfrac{2}{x^{3}}\right)
  3. Multiply out each product term by term.
    \dfrac{dy}{dx}=9x+9-3x-6-\dfrac{27}{x^{2}}-\dfrac{54}{x^{3}}
  4. Collect like terms.
    \dfrac{dy}{dx}=6x+3-\dfrac{27}{x^{2}}-\dfrac{54}{x^{3}}
\dfrac{dy}{dx}=6x+3-\dfrac{27}{x^{2}}-\dfrac{54}{x^{3}}
8
Find \dfrac{dy}{dx} if y=\dfrac{2-3x}{7-x}.
Solution
  1. The function is a quotient. Name the numerator f and the denominator g, and differentiate each.
    \begin{aligned}f&=2-3x\\ f'&=-3\\ g&=7-x\\ g'&=-1\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}: denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared.
    \dfrac{dy}{dx}=\dfrac{(7-x)(-3)-(2-3x)(-1)}{(7-x)^{2}}
  3. Expand the numerator; the x terms cancel.
    \dfrac{dy}{dx}=\dfrac{-21+3x+2-3x}{(7-x)^{2}}=-\dfrac{19}{(7-x)^{2}}
\dfrac{dy}{dx}=-\dfrac{19}{(7-x)^{2}}
9
Find \dfrac{dy}{dx} if y=\dfrac{x^{2}-10x+2}{x^{3}-x}.
Solution
  1. The function is a quotient. Name the numerator f and the denominator g, and differentiate each.
    \begin{aligned}f&=x^{2}-10x+2\\ f'&=2x-10\\ g&=x^{3}-x\\ g'&=3x^{2}-1\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}: denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared.
    \dfrac{dy}{dx}=\dfrac{(x^{3}-x)(2x-10)-(x^{2}-10x+2)(3x^{2}-1)}{(x^{3}-x)^{2}}
  3. Expand the two products in the numerator separately.
    \begin{aligned}(x^{3}-x)(2x-10)&=2x^{4}-10x^{3}-2x^{2}+10x\\ (x^{2}-10x+2)(3x^{2}-1)&=3x^{4}-30x^{3}+5x^{2}+10x-2\end{aligned}
  4. Subtract the second product from the first and collect like terms.
    \dfrac{dy}{dx}=\dfrac{-x^{4}+20x^{3}-7x^{2}+2}{(x^{3}-x)^{2}}
\dfrac{dy}{dx}=\dfrac{-x^{4}+20x^{3}-7x^{2}+2}{(x^{3}-x)^{2}}
10
Find \dfrac{dy}{dx} if y=\dfrac{x^{4}+2x^{3}-1}{x^{2}}.
Solution
  1. The function is a quotient. Name the numerator f and the denominator g, and differentiate each.
    \begin{aligned}f&=x^{4}+2x^{3}-1\\ f'&=4x^{3}+6x^{2}\\ g&=x^{2}\\ g'&=2x\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}: denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared.
    \dfrac{dy}{dx}=\dfrac{x^{2}(4x^{3}+6x^{2})-(x^{4}+2x^{3}-1)(2x)}{(x^{2})^{2}}
  3. Expand the numerator and collect like terms.
    \dfrac{dy}{dx}=\dfrac{4x^{5}+6x^{4}-2x^{5}-4x^{4}+2x}{x^{4}}=\dfrac{2x^{5}+2x^{4}+2x}{x^{4}}
  4. Divide each term of the numerator by x^{4}.
    \dfrac{dy}{dx}=2x+2+\dfrac{2}{x^{3}}
\dfrac{dy}{dx}=2x+2+\dfrac{2}{x^{3}}
11
Find \dfrac{dy}{dx} if y=\dfrac{10}{(x^{3}-10)^{9}}.
Solution
  1. Quotient with numerator f=10, a constant. The derivative of the denominator needs the power rule for functions, \dfrac{d}{dx}[g(x)]^{n}=n[g(x)]^{n-1}g'(x) (Section 2.11.1), with inner function x^{3}-10.
    \begin{aligned}f&=10\\ f'&=0\\ g&=(x^{3}-10)^{9}\\ g'&=9(x^{3}-10)^{8}(3x^{2})=27x^{2}(x^{3}-10)^{8}\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}: denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared.
    \dfrac{dy}{dx}=\dfrac{(x^{3}-10)^{9}(0)-10\cdot27x^{2}(x^{3}-10)^{8}}{\left[(x^{3}-10)^{9}\right]^{2}}
  3. Simplify: \left[(x^{3}-10)^{9}\right]^{2}=(x^{3}-10)^{18}, then cancel (x^{3}-10)^{8} from top and bottom.
    \dfrac{dy}{dx}=\dfrac{-270x^{2}(x^{3}-10)^{8}}{(x^{3}-10)^{18}}=-\dfrac{270x^{2}}{(x^{3}-10)^{10}}
\dfrac{dy}{dx}=-\dfrac{270x^{2}}{(x^{3}-10)^{10}}

Note. The derivative of (x^{3}-10)^{9} needs the power rule for functions, which the book introduces later in Section 2.11.1.

12
Find \dfrac{dy}{dx} if y=\dfrac{(x^{2}+1)^{2}}{3x-2}.
Solution
  1. The function is a quotient. Expand the numerator so that its derivative needs only the power rule.
    \begin{aligned}f&=(x^{2}+1)^{2}=x^{4}+2x^{2}+1\\ f'&=4x^{3}+4x=4x(x^{2}+1)\\ g&=3x-2\\ g'&=3\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}: denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared.
    \dfrac{dy}{dx}=\dfrac{(3x-2)\cdot4x(x^{2}+1)-(x^{2}+1)^{2}(3)}{(3x-2)^{2}}
  3. Take out the common factor (x^{2}+1) from both terms of the numerator.
    \begin{aligned}\dfrac{dy}{dx}&=\dfrac{(x^{2}+1)\left[4x(3x-2)-3(x^{2}+1)\right]}{(3x-2)^{2}}\\ &=\dfrac{(x^{2}+1)(12x^{2}-8x-3x^{2}-3)}{(3x-2)^{2}}\end{aligned}
  4. Collect like terms in the bracket.
    \dfrac{dy}{dx}=\dfrac{(x^{2}+1)(9x^{2}-8x-3)}{(3x-2)^{2}}
\dfrac{dy}{dx}=\dfrac{(x^{2}+1)(9x^{2}-8x-3)}{(3x-2)^{2}}
13
Find \dfrac{dy}{dx} if y=\dfrac{(x+1)^{2}}{(x-1)^{2}}.
Solution
  1. The function is a quotient. Expand both squares so that the power rule gives their derivatives.
    \begin{aligned}f&=(x+1)^{2}=x^{2}+2x+1\\ f'&=2x+2=2(x+1)\\ g&=(x-1)^{2}=x^{2}-2x+1\\ g'&=2x-2=2(x-1)\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}: denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared. Here g^{2}=(x-1)^{4}.
    \dfrac{dy}{dx}=\dfrac{(x-1)^{2}\cdot2(x+1)-(x+1)^{2}\cdot2(x-1)}{(x-1)^{4}}
  3. Take out the common factor 2(x-1)(x+1) from the numerator.
    \begin{aligned}\dfrac{dy}{dx}&=\dfrac{2(x-1)(x+1)\left[(x-1)-(x+1)\right]}{(x-1)^{4}}\\ &=\dfrac{2(x-1)(x+1)(-2)}{(x-1)^{4}}\end{aligned}
  4. Cancel one factor (x-1).
    \dfrac{dy}{dx}=-\dfrac{4(x+1)}{(x-1)^{3}}
\dfrac{dy}{dx}=-\dfrac{4(x+1)}{(x-1)^{3}}
14
Find the slope of the tangent line to the curve at the point whose abscissa is given: y=\dfrac{4x-1}{x}, x=-1.
Solution
  1. The function is a quotient. Name the numerator f and the denominator g, and differentiate each.
    \begin{aligned}f&=4x-1\\ f'&=4\\ g&=x\\ g'&=1\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}: denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared.
    \dfrac{dy}{dx}=\dfrac{x(4)-(4x-1)(1)}{x^{2}}=\dfrac{4x-4x+1}{x^{2}}=\dfrac{1}{x^{2}}
  3. The slope of the tangent at x=-1 is the value of the derivative there.
    m=\dfrac{1}{(-1)^{2}}=1
  4. At x=-1, y=\dfrac{4(-1)-1}{-1}=5, so the point of contact is (-1,5) and the tangent line is y-5=1(x+1), that is y=x+6.
The curve y=4-\dfrac1x with its tangent at (-1,5). The dotted lines are the asymptotes x=0 and y=4.
m=1
15
Find the slope of the tangent line to the curve at the point whose abscissa is given: y=\dfrac{54}{x^{2}+1}, x=2.
Solution
  1. The function is a quotient. Name the numerator f and the denominator g, and differentiate each.
    \begin{aligned}f&=54\\ f'&=0\\ g&=x^{2}+1\\ g'&=2x\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}: denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared.
    \dfrac{dy}{dx}=\dfrac{(x^{2}+1)(0)-54(2x)}{(x^{2}+1)^{2}}=-\dfrac{108x}{(x^{2}+1)^{2}}
  3. Substitute x=2.
    m=-\dfrac{108(2)}{(2^{2}+1)^{2}}=-\dfrac{216}{25}=-8.64
  4. At x=2, y=\dfrac{54}{5}=10.8, so the tangent line is y-10.8=-8.64(x-2).
The curve falls steeply at x=2, so the slope of the tangent is large and negative.
m=-\dfrac{216}{25}=-8.64
16
Find the slope of the tangent line to the curve at the point whose abscissa is given: y=\dfrac{2x+5}{x+2}, x=1.
Solution
  1. The function is a quotient. Name the numerator f and the denominator g, and differentiate each.
    \begin{aligned}f&=2x+5\\ f'&=2\\ g&=x+2\\ g'&=1\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}: denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared.
    \begin{aligned}\dfrac{dy}{dx}&=\dfrac{(x+2)(2)-(2x+5)(1)}{(x+2)^{2}}\\ &=\dfrac{2x+4-2x-5}{(x+2)^{2}}=-\dfrac{1}{(x+2)^{2}}\end{aligned}
  3. Substitute x=1.
    m=-\dfrac{1}{(1+2)^{2}}=-\dfrac{1}{9}
  4. At x=1, y=\dfrac{7}{3}, so the tangent line is y-\dfrac{7}{3}=-\dfrac{1}{9}(x-1).
Near x=1 the curve y=2+\dfrac{1}{x+2} is almost flat, so the slope is small and negative. The dotted line is the asymptote x=-2.
m=-\dfrac{1}{9}
17
Find the slope of the tangent line to the curve at the point whose abscissa is given: y=(2\sqrt{x}+1)(x^{3}-6), x=0.
Solution
  1. The function is a product. Name the two factors f and g, and differentiate each. Write \sqrt{x}=x^{\frac12}.
    \begin{aligned}f&=2\sqrt{x}+1\\ f'&=2\cdot\dfrac{1}{2}x^{-\frac{1}{2}}=\dfrac{1}{\sqrt{x}}\\ g&=x^{3}-6\\ g'&=3x^{2}\end{aligned}
  2. Apply the product rule \dfrac{dy}{dx}=fg'+gf': first function times derivative of the second, plus second function times derivative of the first.
    \dfrac{dy}{dx}=(2\sqrt{x}+1)(3x^{2})+(x^{3}-6)\dfrac{1}{\sqrt{x}}
  3. Simplify, using \sqrt{x}\cdot x^{2}=x^{\frac52} and \dfrac{x^{3}}{\sqrt{x}}=x^{\frac52}.
    \dfrac{dy}{dx}=6x^{\frac{5}{2}}+3x^{2}+x^{\frac{5}{2}}-\dfrac{6}{\sqrt{x}}=7x^{\frac{5}{2}}+3x^{2}-\dfrac{6}{\sqrt{x}}
  4. Now try x=0. The last term -\dfrac{6}{\sqrt{x}} has zero in its denominator, so \dfrac{dy}{dx} has no value at x=0. As x\to0^{+} this term grows without bound in the negative direction.
    \lim_{x\to0^{+}}\dfrac{dy}{dx}=\lim_{x\to0^{+}}\left(7x^{\frac{5}{2}}+3x^{2}-\dfrac{6}{\sqrt{x}}\right)=-\infty
  5. So the slope of the tangent at x=0 does not exist. The curve is defined only for x\ge0 and starts at (0,-6); it leaves this point going straight down, so the tangent there is the vertical line x=0.
At its starting point (0,-6) the curve is vertical, so the tangent is the line x=0 and its slope does not exist.
m\ \text{does not exist at } x=0\ (\text{vertical tangent})

Exercise 2.6p. 65

Questions 1 to 10 use the derivatives of Section 2.9: \dfrac{d}{dx}\sin x=\cos x, \dfrac{d}{dx}\cos x=-\sin x, \dfrac{d}{dx}\tan x=\sec^{2}x, \dfrac{d}{dx}\cot x=-\operatorname{cosec}^{2}x, \dfrac{d}{dx}\sec x=\sec x\tan x, together with the product and quotient rules. Questions 11 to 16 use the inverse trigonometric derivatives of Section 2.10, with the inner function differentiated as in Examples 25 to 27.

1
Find the derivative of the given function w.r.t. x: y=x^{2}-\cos x.
Solution
  1. Differentiate term by term, using \dfrac{d}{dx}\cos x=-\sin x.
    \dfrac{dy}{dx}=\dfrac{d}{dx}x^{2}-\dfrac{d}{dx}\cos x=2x-(-\sin x)
  2. Two minus signs make a plus.
    \dfrac{dy}{dx}=2x+\sin x
\dfrac{dy}{dx}=2x+\sin x
2
Find the derivative of the given function w.r.t. x: y=4x^{3}+x+\sin x.
Solution
  1. Differentiate each term: the power rule for 4x^{3} and x, and \dfrac{d}{dx}\sin x=\cos x.
    \dfrac{dy}{dx}=4(3x^{2})+1+\cos x
  2. Simplify.
    \dfrac{dy}{dx}=12x^{2}+1+\cos x
\dfrac{dy}{dx}=12x^{2}+1+\cos x
3
Find the derivative of the given function w.r.t. x: y=3\cos x-5\cot x.
Solution
  1. Use \dfrac{d}{dx}\cos x=-\sin x and \dfrac{d}{dx}\cot x=-\operatorname{cosec}^{2}x, keeping the constant multipliers 3 and 5.
    \dfrac{dy}{dx}=3(-\sin x)-5\left(-\operatorname{cosec}^{2}x\right)
  2. Simplify the signs.
    \dfrac{dy}{dx}=-3\sin x+5\operatorname{cosec}^{2}x
\dfrac{dy}{dx}=5\operatorname{cosec}^{2}x-3\sin x
4
Find the derivative of the given function w.r.t. x: y=\sin x\cos x.
Solution
  1. The function is a product. Name the two factors f and g, and differentiate each.
    \begin{aligned}f&=\sin x\\ f'&=\cos x\\ g&=\cos x\\ g'&=-\sin x\end{aligned}
  2. Apply the product rule \dfrac{dy}{dx}=fg'+gf'.
    \dfrac{dy}{dx}=\sin x(-\sin x)+\cos x(\cos x)=\cos^{2}x-\sin^{2}x
  3. Use the double angle identity \cos2x=\cos^{2}x-\sin^{2}x.
    \dfrac{dy}{dx}=\cos 2x
\dfrac{dy}{dx}=\cos^{2}x-\sin^{2}x=\cos 2x
5
Find the derivative of the given function w.r.t. x: y=(x^{2}+\sin x)\sec x.
Solution
  1. The function is a product. Name the two factors f and g, and differentiate each. Use \dfrac{d}{dx}\sec x=\sec x\tan x.
    \begin{aligned}f&=x^{2}+\sin x\\ f'&=2x+\cos x\\ g&=\sec x\\ g'&=\sec x\tan x\end{aligned}
  2. Apply the product rule \dfrac{dy}{dx}=fg'+gf'.
    \dfrac{dy}{dx}=(x^{2}+\sin x)\sec x\tan x+\sec x\,(2x+\cos x)
  3. Multiply out, using \sin x\sec x=\dfrac{\sin x}{\cos x}=\tan x and \sec x\cos x=1.
    \dfrac{dy}{dx}=x^{2}\sec x\tan x+\tan^{2}x+2x\sec x+1
  4. Use the identity 1+\tan^{2}x=\sec^{2}x.
    \dfrac{dy}{dx}=x^{2}\sec x\tan x+2x\sec x+\sec^{2}x
\dfrac{dy}{dx}=x^{2}\sec x\tan x+2x\sec x+\sec^{2}x
6
Find the derivative of the given function w.r.t. x: y=\dfrac{5-\cos x}{5+\sin x}.
Solution
  1. The function is a quotient. Name the numerator f and the denominator g, and differentiate each.
    \begin{aligned}f&=5-\cos x\\ f'&=\sin x\\ g&=5+\sin x\\ g'&=\cos x\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}.
    \dfrac{dy}{dx}=\dfrac{(5+\sin x)\sin x-(5-\cos x)\cos x}{(5+\sin x)^{2}}
  3. Expand the numerator and use \sin^{2}x+\cos^{2}x=1.
    \begin{aligned}\dfrac{dy}{dx}&=\dfrac{5\sin x+\sin^{2}x-5\cos x+\cos^{2}x}{(5+\sin x)^{2}}\\ &=\dfrac{5\sin x-5\cos x+1}{(5+\sin x)^{2}}\end{aligned}
\dfrac{dy}{dx}=\dfrac{5\sin x-5\cos x+1}{(5+\sin x)^{2}}
7
Find the derivative of the given function w.r.t. x: y=\dfrac{\sec x}{1+\tan x}.
Solution
  1. The function is a quotient. Name the numerator f and the denominator g, and differentiate each.
    \begin{aligned}f&=\sec x\\ f'&=\sec x\tan x\\ g&=1+\tan x\\ g'&=\sec^{2}x\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}.
    \dfrac{dy}{dx}=\dfrac{(1+\tan x)\sec x\tan x-\sec x\cdot\sec^{2}x}{(1+\tan x)^{2}}
  3. Take out the common factor \sec x and replace \sec^{2}x by 1+\tan^{2}x.
    \dfrac{dy}{dx}=\dfrac{\sec x\left(\tan x+\tan^{2}x-1-\tan^{2}x\right)}{(1+\tan x)^{2}}
  4. The \tan^{2}x terms cancel.
    \dfrac{dy}{dx}=\dfrac{\sec x(\tan x-1)}{(1+\tan x)^{2}}
\dfrac{dy}{dx}=\dfrac{\sec x(\tan x-1)}{(1+\tan x)^{2}}
8
Find the derivative of the given function w.r.t. x: y=\dfrac{\sin x}{x^{2}+\sin x}.
Solution
  1. The function is a quotient. Name the numerator f and the denominator g, and differentiate each.
    \begin{aligned}f&=\sin x\\ f'&=\cos x\\ g&=x^{2}+\sin x\\ g'&=2x+\cos x\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}.
    \dfrac{dy}{dx}=\dfrac{(x^{2}+\sin x)\cos x-\sin x(2x+\cos x)}{(x^{2}+\sin x)^{2}}
  3. Expand the numerator; the terms \sin x\cos x cancel.
    \begin{aligned}\dfrac{dy}{dx}&=\dfrac{\begin{aligned}&x^{2}\cos x+\sin x\cos x\\ &-2x\sin x-\sin x\cos x\end{aligned}}{(x^{2}+\sin x)^{2}}\\ &=\dfrac{x^{2}\cos x-2x\sin x}{(x^{2}+\sin x)^{2}}\end{aligned}
  4. Take out the common factor x (optional).
    \dfrac{dy}{dx}=\dfrac{x(x\cos x-2\sin x)}{(x^{2}+\sin x)^{2}}
\dfrac{dy}{dx}=\dfrac{x^{2}\cos x-2x\sin x}{(x^{2}+\sin x)^{2}}
9
Find the derivative of the given function w.r.t. x: y=\dfrac{\cot x}{x+1}.
Solution
  1. The function is a quotient. Name the numerator f and the denominator g, and differentiate each. Use \dfrac{d}{dx}\cot x=-\operatorname{cosec}^{2}x.
    \begin{aligned}f&=\cot x\\ f'&=-\operatorname{cosec}^{2}x\\ g&=x+1\\ g'&=1\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}.
    \dfrac{dy}{dx}=\dfrac{(x+1)\left(-\operatorname{cosec}^{2}x\right)-\cot x\,(1)}{(x+1)^{2}}
  3. Take the minus sign out of the numerator.
    \dfrac{dy}{dx}=-\dfrac{(x+1)\operatorname{cosec}^{2}x+\cot x}{(x+1)^{2}}
\dfrac{dy}{dx}=-\dfrac{(x+1)\operatorname{cosec}^{2}x+\cot x}{(x+1)^{2}}
10
Find the derivative of the given function w.r.t. x: y=(1+\cos x)(x-\sin x).
Solution
  1. The function is a product. Name the two factors f and g, and differentiate each.
    \begin{aligned}f&=1+\cos x\\ f'&=-\sin x\\ g&=x-\sin x\\ g'&=1-\cos x\end{aligned}
  2. Apply the product rule \dfrac{dy}{dx}=fg'+gf'.
    \dfrac{dy}{dx}=(1+\cos x)(1-\cos x)+(x-\sin x)(-\sin x)
  3. Use (1+\cos x)(1-\cos x)=1-\cos^{2}x=\sin^{2}x and expand the second product.
    \dfrac{dy}{dx}=\sin^{2}x-x\sin x+\sin^{2}x
  4. Collect like terms and factorise.
    \dfrac{dy}{dx}=2\sin^{2}x-x\sin x=\sin x\,(2\sin x-x)
\dfrac{dy}{dx}=2\sin^{2}x-x\sin x
11
Find the derivative of the given function w.r.t. x: y=\sin^{-1}(5x-1).
Solution
  1. Use \dfrac{d}{dx}\sin^{-1}u=\dfrac{1}{\sqrt{1-u^{2}}}\cdot\dfrac{du}{dx} (as in Example 25) with the inner function u=5x-1.
    u=5x-1,\quad \dfrac{du}{dx}=5
  2. Substitute.
    \dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-(5x-1)^{2}}}\cdot5=\dfrac{5}{\sqrt{1-(5x-1)^{2}}}
  3. Simplify under the root: 1-(25x^{2}-10x+1)=10x-25x^{2}.
    \dfrac{dy}{dx}=\dfrac{5}{\sqrt{10x-25x^{2}}}
  4. The formula needs -1<5x-1<1, that is 0<x<\tfrac25.
\dfrac{dy}{dx}=\dfrac{5}{\sqrt{1-(5x-1)^{2}}}=\dfrac{5}{\sqrt{10x-25x^{2}}}
12
Find the derivative of the given function w.r.t. x: y=4\cot^{-1}\dfrac{x}{2}.
Solution
  1. Use \dfrac{d}{dx}\cot^{-1}u=-\dfrac{1}{1+u^{2}}\cdot\dfrac{du}{dx} with u=\dfrac{x}{2}, so \dfrac{du}{dx}=\dfrac12. The constant 4 stays as a multiplier.
    \dfrac{dy}{dx}=4\left(-\dfrac{1}{1+\left(\frac{x}{2}\right)^{2}}\right)\cdot\dfrac{1}{2}
  2. Simplify, using 1+\dfrac{x^{2}}{4}=\dfrac{4+x^{2}}{4}.
    \dfrac{dy}{dx}=-\dfrac{2}{1+\frac{x^{2}}{4}}=-\dfrac{8}{4+x^{2}}
\dfrac{dy}{dx}=-\dfrac{8}{4+x^{2}}
13
Find the derivative of the given function w.r.t. x: y=\dfrac{\sin^{-1}x}{\sin x}.
Solution
  1. The function is a quotient. Name the numerator f and the denominator g, and differentiate each. Use \dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^{2}}}.
    \begin{aligned}f&=\sin^{-1}x\\ f'&=\dfrac{1}{\sqrt{1-x^{2}}}\\ g&=\sin x\\ g'&=\cos x\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}.
    \dfrac{dy}{dx}=\dfrac{\sin x\cdot\dfrac{1}{\sqrt{1-x^{2}}}-\sin^{-1}x\cdot\cos x}{\sin^{2}x}
  3. Multiply the top and the bottom by \sqrt{1-x^{2}} to clear the fraction inside the numerator.
    \dfrac{dy}{dx}=\dfrac{\sin x-\sqrt{1-x^{2}}\,\cos x\,\sin^{-1}x}{\sqrt{1-x^{2}}\,\sin^{2}x}
  4. This holds for -1<x<1 with x\neq0 (where \sin x=0).
\dfrac{dy}{dx}=\dfrac{\sin x-\sqrt{1-x^{2}}\,\cos x\,\sin^{-1}x}{\sqrt{1-x^{2}}\,\sin^{2}x}
14
Find the derivative of the given function w.r.t. x: y=\dfrac{\sec^{-1}x}{x}.
Solution
  1. The function is a quotient. Name the numerator f and the denominator g, and differentiate each. Use \dfrac{d}{dx}\sec^{-1}x=\dfrac{1}{x\sqrt{x^{2}-1}} for |x|>1 (Section 2.10).
    \begin{aligned}f&=\sec^{-1}x\\ f'&=\dfrac{1}{x\sqrt{x^{2}-1}}\\ g&=x\\ g'&=1\end{aligned}
  2. Substitute into the quotient rule \dfrac{dy}{dx}=\dfrac{gf'-fg'}{g^{2}}.
    \dfrac{dy}{dx}=\dfrac{x\cdot\dfrac{1}{x\sqrt{x^{2}-1}}-\sec^{-1}x\cdot1}{x^{2}}
  3. Cancel x in the first term of the numerator.
    \dfrac{dy}{dx}=\dfrac{\dfrac{1}{\sqrt{x^{2}-1}}-\sec^{-1}x}{x^{2}}
  4. Multiply the top and the bottom by \sqrt{x^{2}-1}.
    \dfrac{dy}{dx}=\dfrac{1-\sqrt{x^{2}-1}\,\sec^{-1}x}{x^{2}\sqrt{x^{2}-1}}
\dfrac{dy}{dx}=\dfrac{1-\sqrt{x^{2}-1}\,\sec^{-1}x}{x^{2}\sqrt{x^{2}-1}},\quad |x|>1
15
Find the derivative of the given function w.r.t. x: y=x\sin^{-1}x+x\cos^{-1}x.
Solution
  1. Each term is a product, so use the product rule on each. Use \dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^{2}}} and \dfrac{d}{dx}\cos^{-1}x=-\dfrac{1}{\sqrt{1-x^{2}}}.
    \dfrac{dy}{dx}=\left(x\cdot\dfrac{1}{\sqrt{1-x^{2}}}+\sin^{-1}x\right)+\left(x\cdot\dfrac{-1}{\sqrt{1-x^{2}}}+\cos^{-1}x\right)
  2. The two fractions cancel.
    \dfrac{dy}{dx}=\sin^{-1}x+\cos^{-1}x
  3. Use the identity \sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2} for -1\le x\le1.
    \dfrac{dy}{dx}=\dfrac{\pi}{2}
y=x\sin^{-1}x+x\cos^{-1}x is the segment y=\tfrac{\pi}{2}x for -1\le x\le1.
\dfrac{dy}{dx}=\dfrac{\pi}{2}
16
Find the derivative of the given function w.r.t. x: y=\dfrac{1}{\tan^{-1}x^{2}}.
Solution
  1. Here \tan^{-1}x^{2} means \tan^{-1}(x^{2}). Differentiate it first with \dfrac{d}{dx}\tan^{-1}u=\dfrac{1}{1+u^{2}}\cdot\dfrac{du}{dx} and u=x^{2}.
    \dfrac{d}{dx}\tan^{-1}(x^{2})=\dfrac{1}{1+(x^{2})^{2}}\cdot2x=\dfrac{2x}{1+x^{4}}
  2. Quotient rule with f=1, f'=0 and g=\tan^{-1}x^{2}.
    \dfrac{dy}{dx}=\dfrac{\tan^{-1}x^{2}\cdot0-1\cdot\dfrac{2x}{1+x^{4}}}{\left(\tan^{-1}x^{2}\right)^{2}}
  3. Simplify.
    \dfrac{dy}{dx}=-\dfrac{2x}{(1+x^{4})\left(\tan^{-1}x^{2}\right)^{2}}
\dfrac{dy}{dx}=-\dfrac{2x}{(1+x^{4})\left(\tan^{-1}x^{2}\right)^{2}}

Note. \tan^{-1}x^{2} is read as \tan^{-1}(x^{2}), in the same way as \sec^{-1}x^{2} in the book's Example 27.

Exercise 2.7p. 71–72

Practises the chain rule and the power rule for functions (Section 2.11), implicit differentiation (2.12), derivatives of e^{x} and \ln x (2.13, 2.14), parametric functions, differentials (2.15) and approximations (2.16).

1
Find the derivative of the function w.r.t. the variable involved: y=\left(x-\dfrac{1}{x^{2}}\right)^{5}
Solution
  1. The function is a power of a function, y=[g(x)]^{5} with g(x)=x-x^{-2}. Use the power rule for functions, \frac{d}{dx}[g(x)]^{n}=n[g(x)]^{n-1}g'(x).
    \dfrac{dy}{dx}=5\left(x-\dfrac{1}{x^{2}}\right)^{4}\dfrac{d}{dx}\left(x-x^{-2}\right)
  2. Differentiate the inside term by term with the power rule.
    \dfrac{d}{dx}\left(x-x^{-2}\right)=1+2x^{-3}=1+\dfrac{2}{x^{3}}
  3. Put the two pieces together.
    \dfrac{dy}{dx}=5\left(x-\dfrac{1}{x^{2}}\right)^{4}\left(1+\dfrac{2}{x^{3}}\right)
  4. For a single fraction write x-\frac{1}{x^{2}}=\frac{x^{3}-1}{x^{2}} and 1+\frac{2}{x^{3}}=\frac{x^{3}+2}{x^{3}}.
    \dfrac{dy}{dx}=5\cdot\dfrac{(x^{3}-1)^{4}}{x^{8}}\cdot\dfrac{x^{3}+2}{x^{3}}=\dfrac{5(x^{3}-1)^{4}(x^{3}+2)}{x^{11}}
\dfrac{dy}{dx}=5\left(x-\dfrac{1}{x^{2}}\right)^{4}\left(1+\dfrac{2}{x^{3}}\right)=\dfrac{5(x^{3}-1)^{4}(x^{3}+2)}{x^{11}}
2
Find the derivative of the function w.r.t. the variable involved: f(x)=\left(\dfrac{x^{2}-1}{x^{2}+1}\right)^{2}
Solution
  1. Power rule for functions with n=2. The inside function is the quotient \frac{x^{2}-1}{x^{2}+1}.
    f'(x)=2\left(\dfrac{x^{2}-1}{x^{2}+1}\right)\dfrac{d}{dx}\left(\dfrac{x^{2}-1}{x^{2}+1}\right)
  2. Differentiate the inside with the quotient rule, \left(\frac{u}{v}\right)'=\frac{vu'-uv'}{v^{2}}.
    \dfrac{d}{dx}\left(\dfrac{x^{2}-1}{x^{2}+1}\right)=\dfrac{(x^{2}+1)(2x)-(x^{2}-1)(2x)}{(x^{2}+1)^{2}}=\dfrac{4x}{(x^{2}+1)^{2}}
  3. Multiply the two factors.
    f'(x)=2\cdot\dfrac{x^{2}-1}{x^{2}+1}\cdot\dfrac{4x}{(x^{2}+1)^{2}}=\dfrac{8x(x^{2}-1)}{(x^{2}+1)^{3}}
f'(x)=\dfrac{8x(x^{2}-1)}{(x^{2}+1)^{3}}
3
Find the derivative of the function w.r.t. the variable involved: y=(3x-1)^{4}(-2x+9)^{5}
Solution
  1. Product rule, \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}, with u=(3x-1)^{4} and v=(-2x+9)^{5}.
    \dfrac{dy}{dx}=(3x-1)^{4}\dfrac{d}{dx}(-2x+9)^{5}+(-2x+9)^{5}\dfrac{d}{dx}(3x-1)^{4}
  2. Power rule for functions on each bracket. The inside derivatives are -2 and 3.
    \dfrac{d}{dx}(-2x+9)^{5}=5(-2x+9)^{4}(-2)=-10(-2x+9)^{4},\qquad \dfrac{d}{dx}(3x-1)^{4}=4(3x-1)^{3}(3)=12(3x-1)^{3}
  3. Substitute.
    \dfrac{dy}{dx}=-10(3x-1)^{4}(-2x+9)^{4}+12(3x-1)^{3}(-2x+9)^{5}
  4. Take out the common factor 2(3x-1)^{3}(-2x+9)^{4}. Inside the bracket, -5(3x-1)+6(-2x+9)=-15x+5-12x+54=59-27x.
    \dfrac{dy}{dx}=2(3x-1)^{3}(-2x+9)^{4}\left[-5(3x-1)+6(-2x+9)\right]=2(3x-1)^{3}(-2x+9)^{4}(59-27x)
\dfrac{dy}{dx}=2(3x-1)^{3}(-2x+9)^{4}(59-27x)
4
Find the derivative of the function w.r.t. the variable involved: f(\theta)=(2\theta+1)^{3}\tan^{2}\theta
Solution
  1. Product rule with u=(2\theta+1)^{3} and v=\tan^{2}\theta.
    f'(\theta)=(2\theta+1)^{3}\dfrac{d}{d\theta}\left(\tan^{2}\theta\right)+\tan^{2}\theta\,\dfrac{d}{d\theta}(2\theta+1)^{3}
  2. \tan^{2}\theta=(\tan\theta)^{2}, so use the power rule for functions (as in Example 30a) and \frac{d}{d\theta}\tan\theta=\sec^{2}\theta.
    \dfrac{d}{d\theta}\tan^{2}\theta=2\tan\theta\sec^{2}\theta,\qquad \dfrac{d}{d\theta}(2\theta+1)^{3}=3(2\theta+1)^{2}(2)=6(2\theta+1)^{2}
  3. Substitute.
    f'(\theta)=2(2\theta+1)^{3}\tan\theta\sec^{2}\theta+6(2\theta+1)^{2}\tan^{2}\theta
  4. Take out the common factor 2(2\theta+1)^{2}\tan\theta.
    f'(\theta)=2(2\theta+1)^{2}\tan\theta\left[(2\theta+1)\sec^{2}\theta+3\tan\theta\right]
f'(\theta)=2(2\theta+1)^{2}\tan\theta\left[(2\theta+1)\sec^{2}\theta+3\tan\theta\right]
5
Find the derivative of the function w.r.t. the variable involved: y=\sin 2x\cos 3x
Solution
  1. Product rule with u=\sin 2x and v=\cos 3x.
    \dfrac{dy}{dx}=\sin 2x\,\dfrac{d}{dx}(\cos 3x)+\cos 3x\,\dfrac{d}{dx}(\sin 2x)
  2. Chain rule: the derivative of the angle (3 or 2) multiplies the result.
    \dfrac{d}{dx}\cos 3x=-3\sin 3x,\qquad \dfrac{d}{dx}\sin 2x=2\cos 2x
  3. Substitute and arrange.
    \dfrac{dy}{dx}=-3\sin 2x\sin 3x+2\cos 2x\cos 3x=2\cos 2x\cos 3x-3\sin 2x\sin 3x
\dfrac{dy}{dx}=2\cos 2x\cos 3x-3\sin 2x\sin 3x
6
Find the derivative of the function w.r.t. the variable involved: f(x)=(\sec 4x+\tan 2x)^{5}
Solution
  1. Power rule for functions with n=5.
    f'(x)=5(\sec 4x+\tan 2x)^{4}\dfrac{d}{dx}(\sec 4x+\tan 2x)
  2. Chain rule on each term: \frac{d}{dx}\sec u=\sec u\tan u\,\frac{du}{dx} and \frac{d}{dx}\tan u=\sec^{2}u\,\frac{du}{dx}.
    \dfrac{d}{dx}(\sec 4x+\tan 2x)=4\sec 4x\tan 4x+2\sec^{2}2x
  3. Substitute and take out the common factor 2.
    f'(x)=5(\sec 4x+\tan 2x)^{4}(4\sec 4x\tan 4x+2\sec^{2}2x)=10(\sec 4x+\tan 2x)^{4}(2\sec 4x\tan 4x+\sec^{2}2x)
f'(x)=10(\sec 4x+\tan 2x)^{4}(2\sec 4x\tan 4x+\sec^{2}2x)
7
Find the derivative of the function w.r.t. the variable involved: h(t)=\dfrac{t+\sin 4t}{10+\cos 3t}
Solution
  1. Quotient rule with u=t+\sin 4t and v=10+\cos 3t: h'(t)=\dfrac{vu'-uv'}{v^{2}}.
    h'(t)=\dfrac{(10+\cos 3t)\dfrac{d}{dt}(t+\sin 4t)-(t+\sin 4t)\dfrac{d}{dt}(10+\cos 3t)}{(10+\cos 3t)^{2}}
  2. Differentiate the top and bottom with the chain rule.
    \dfrac{d}{dt}(t+\sin 4t)=1+4\cos 4t,\qquad \dfrac{d}{dt}(10+\cos 3t)=-3\sin 3t
  3. Substitute. The two minus signs make a plus.
    h'(t)=\dfrac{(10+\cos 3t)(1+4\cos 4t)+3\sin 3t\,(t+\sin 4t)}{(10+\cos 3t)^{2}}
h'(t)=\dfrac{(10+\cos 3t)(1+4\cos 4t)+3\sin 3t\,(t+\sin 4t)}{(10+\cos 3t)^{2}}
8
Find the derivative of the function w.r.t. the variable involved: f(x)=\tan\left(\cos\dfrac{x}{2}\right)
Solution
  1. This is a function of a function of a function: \tan of \cos of \frac{x}{2}. Apply the chain rule from the outside in. First, \frac{d}{dx}\tan u=\sec^{2}u\frac{du}{dx} with u=\cos\frac{x}{2}.
    f'(x)=\sec^{2}\left(\cos\dfrac{x}{2}\right)\dfrac{d}{dx}\left(\cos\dfrac{x}{2}\right)
  2. Next, \frac{d}{dx}\cos\frac{x}{2}=-\sin\frac{x}{2}\cdot\frac{d}{dx}\left(\frac{x}{2}\right) and \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{1}{2}.
    \dfrac{d}{dx}\left(\cos\dfrac{x}{2}\right)=-\dfrac{1}{2}\sin\dfrac{x}{2}
  3. Multiply.
    f'(x)=-\dfrac{1}{2}\sin\dfrac{x}{2}\sec^{2}\left(\cos\dfrac{x}{2}\right)
f'(x)=-\dfrac{1}{2}\sin\dfrac{x}{2}\sec^{2}\left(\cos\dfrac{x}{2}\right)
9
Use implicit differentiation to find \frac{dy}{dx}: 4x^{2}+y^{2}=8
Solution
  1. Differentiate both sides w.r.t. x. The derivative of the constant 8 is 0.
    8x+2y\dfrac{dy}{dx}=0
  2. Solve for \frac{dy}{dx}.
    \dfrac{dy}{dx}=-\dfrac{8x}{2y}=-\dfrac{4x}{y}
\dfrac{dy}{dx}=-\dfrac{4x}{y}
10
Use implicit differentiation to find \frac{dy}{dx}: x+xy-y^{2}-20=0
Solution
  1. Differentiate each term w.r.t. x. Use the product rule on xy: \frac{d}{dx}(xy)=x\frac{dy}{dx}+y.
    1+x\dfrac{dy}{dx}+y-2y\dfrac{dy}{dx}=0
  2. Keep the \frac{dy}{dx} terms on the left and move the rest to the right.
    (x-2y)\dfrac{dy}{dx}=-(1+y)
  3. Divide by x-2y.
    \dfrac{dy}{dx}=-\dfrac{1+y}{x-2y}=\dfrac{1+y}{2y-x}
\dfrac{dy}{dx}=\dfrac{1+y}{2y-x}
11
Use implicit differentiation to find \frac{dy}{dx}: y^{4}-y^{2}=10x-3
Solution
  1. Differentiate both sides w.r.t. x. Each y term gets a factor \frac{dy}{dx}.
    4y^{3}\dfrac{dy}{dx}-2y\dfrac{dy}{dx}=10
  2. Take \frac{dy}{dx} common.
    (4y^{3}-2y)\dfrac{dy}{dx}=10
  3. Divide and cancel the common factor 2.
    \dfrac{dy}{dx}=\dfrac{10}{4y^{3}-2y}=\dfrac{5}{2y^{3}-y}=\dfrac{5}{y(2y^{2}-1)}
\dfrac{dy}{dx}=\dfrac{5}{2y^{3}-y}
12
Use implicit differentiation to find \frac{dy}{dx}: x^{3}y^{2}=2x^{2}+y^{2}
Solution
  1. Differentiate both sides. The left side is a product, so use the product rule.
    x^{3}\cdot 2y\dfrac{dy}{dx}+y^{2}\cdot 3x^{2}=4x+2y\dfrac{dy}{dx}
  2. Collect the \frac{dy}{dx} terms on the left.
    \left(2x^{3}y-2y\right)\dfrac{dy}{dx}=4x-3x^{2}y^{2}
  3. Divide by 2x^{3}y-2y=2y(x^{3}-1).
    \dfrac{dy}{dx}=\dfrac{4x-3x^{2}y^{2}}{2x^{3}y-2y}=\dfrac{x(4-3xy^{2})}{2y(x^{3}-1)}
\dfrac{dy}{dx}=\dfrac{x(4-3xy^{2})}{2y(x^{3}-1)}
13
Use implicit differentiation to find \frac{dy}{dx}: xy=\sin x+y
Solution
  1. Differentiate both sides; product rule on xy.
    x\dfrac{dy}{dx}+y=\cos x+\dfrac{dy}{dx}
  2. Collect the \frac{dy}{dx} terms on the left.
    (x-1)\dfrac{dy}{dx}=\cos x-y
  3. Divide by x-1.
    \dfrac{dy}{dx}=\dfrac{\cos x-y}{x-1}
\dfrac{dy}{dx}=\dfrac{\cos x-y}{x-1}
14
Use implicit differentiation to find \frac{dy}{dx}: x+y=\cos xy
Solution
  1. Differentiate both sides. For \cos(xy) use the chain rule, then the product rule on xy.
    1+\dfrac{dy}{dx}=-\sin(xy)\left(x\dfrac{dy}{dx}+y\right)
  2. Open the bracket.
    1+\dfrac{dy}{dx}=-x\sin(xy)\dfrac{dy}{dx}-y\sin(xy)
  3. Collect the \frac{dy}{dx} terms on the left.
    \left(1+x\sin(xy)\right)\dfrac{dy}{dx}=-1-y\sin(xy)
  4. Divide.
    \dfrac{dy}{dx}=-\dfrac{1+y\sin(xy)}{1+x\sin(xy)}
\dfrac{dy}{dx}=-\dfrac{1+y\sin(xy)}{1+x\sin(xy)}
15
Use implicit differentiation to find \frac{dy}{dx}: x\sin y-y\cos x=1
Solution
  1. Differentiate both sides. Both terms are products, so use the product rule; also \frac{d}{dx}\sin y=\cos y\frac{dy}{dx}.
    \left(x\cos y\dfrac{dy}{dx}+\sin y\right)-\left(y(-\sin x)+\cos x\dfrac{dy}{dx}\right)=0
  2. Remove the brackets and collect the \frac{dy}{dx} terms.
    (x\cos y-\cos x)\dfrac{dy}{dx}=-\sin y-y\sin x
  3. Divide by x\cos y-\cos x.
    \dfrac{dy}{dx}=-\dfrac{\sin y+y\sin x}{x\cos y-\cos x}=\dfrac{\sin y+y\sin x}{\cos x-x\cos y}
\dfrac{dy}{dx}=\dfrac{\sin y+y\sin x}{\cos x-x\cos y}
16
Use implicit differentiation to find \frac{dy}{dx}: \sin y=y\cos 2x
Solution
  1. Differentiate both sides. Left: chain rule. Right: product rule, with \frac{d}{dx}\cos 2x=-2\sin 2x.
    \cos y\dfrac{dy}{dx}=y(-2\sin 2x)+\cos 2x\dfrac{dy}{dx}
  2. Collect the \frac{dy}{dx} terms on the left.
    (\cos y-\cos 2x)\dfrac{dy}{dx}=-2y\sin 2x
  3. Divide.
    \dfrac{dy}{dx}=\dfrac{-2y\sin 2x}{\cos y-\cos 2x}=\dfrac{2y\sin 2x}{\cos 2x-\cos y}
\dfrac{dy}{dx}=\dfrac{2y\sin 2x}{\cos 2x-\cos y}
17
Find \frac{dy}{dx}: y=x^{3}e^{5x}
Solution
  1. Product rule with u=x^{3} and v=e^{5x}. By the chain rule \frac{d}{dx}e^{5x}=5e^{5x} (as in Example 32).
    \dfrac{dy}{dx}=x^{3}\dfrac{d}{dx}e^{5x}+e^{5x}\dfrac{d}{dx}x^{3}
  2. Differentiate each factor.
    \dfrac{dy}{dx}=5x^{3}e^{5x}+3x^{2}e^{5x}
  3. Take out the common factor x^{2}e^{5x}.
    \dfrac{dy}{dx}=x^{2}e^{5x}(5x+3)
\dfrac{dy}{dx}=x^{2}e^{5x}(5x+3)
18
Find \frac{dy}{dx}: y=e^{4x}(1+\ln x)
Solution
  1. Product rule with u=e^{4x} and v=1+\ln x. Use \frac{d}{dx}e^{4x}=4e^{4x} and \frac{d}{dx}\ln x=\frac{1}{x}.
    \dfrac{dy}{dx}=e^{4x}\dfrac{d}{dx}(1+\ln x)+(1+\ln x)\dfrac{d}{dx}e^{4x}
  2. Differentiate each factor.
    \dfrac{dy}{dx}=e^{4x}\cdot\dfrac{1}{x}+4e^{4x}(1+\ln x)
  3. Take out e^{4x} and write the bracket as one fraction.
    \dfrac{dy}{dx}=e^{4x}\left(\dfrac{1}{x}+4+4\ln x\right)=\dfrac{e^{4x}(1+4x+4x\ln x)}{x}
\dfrac{dy}{dx}=\dfrac{e^{4x}(1+4x+4x\ln x)}{x}
19
Find \frac{dy}{dx}: y=\dfrac{e^{2x}}{e^{-2x}+1}
Solution
  1. Quotient rule with u=e^{2x} and v=e^{-2x}+1, so u'=2e^{2x} and v'=-2e^{-2x}.
    \dfrac{dy}{dx}=\dfrac{(e^{-2x}+1)(2e^{2x})-e^{2x}(-2e^{-2x})}{(e^{-2x}+1)^{2}}
  2. Simplify the numerator using e^{-2x}\cdot e^{2x}=e^{0}=1.
    (e^{-2x}+1)(2e^{2x})+2e^{2x}e^{-2x}=2+2e^{2x}+2=2e^{2x}+4
  3. So the derivative is
    \dfrac{dy}{dx}=\dfrac{2e^{2x}+4}{(e^{-2x}+1)^{2}}=\dfrac{2(e^{2x}+2)}{(e^{-2x}+1)^{2}}
  4. A tidier form: multiply top and bottom by e^{4x}, because e^{4x}(e^{-2x}+1)^{2}=(1+e^{2x})^{2}.
    \dfrac{dy}{dx}=\dfrac{2e^{4x}(e^{2x}+2)}{(e^{2x}+1)^{2}}
\dfrac{dy}{dx}=\dfrac{2(e^{2x}+2)}{(e^{-2x}+1)^{2}}=\dfrac{2e^{4x}(e^{2x}+2)}{(e^{2x}+1)^{2}}
20
Find \frac{dy}{dx}: y=\ln(e^{x}+e^{-x})
Solution
  1. Use \frac{d}{dx}\ln u=\frac{1}{u}\frac{du}{dx} (Section 2.14 with the chain rule), where u=e^{x}+e^{-x}.
    \dfrac{dy}{dx}=\dfrac{1}{e^{x}+e^{-x}}\cdot\dfrac{d}{dx}\left(e^{x}+e^{-x}\right)
  2. Since \frac{d}{dx}e^{-x}=-e^{-x}:
    \dfrac{dy}{dx}=\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}}
  3. Multiplying top and bottom by e^{x} gives an equivalent form.
    \dfrac{dy}{dx}=\dfrac{e^{2x}-1}{e^{2x}+1}
\dfrac{dy}{dx}=\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}}
21
Find \frac{dy}{dx}: y=\ln\left(x+\sqrt{x^{2}+1}\right)
Solution
  1. Chain rule for \ln u with u=x+\sqrt{x^{2}+1}.
    \dfrac{dy}{dx}=\dfrac{1}{x+\sqrt{x^{2}+1}}\cdot\dfrac{d}{dx}\left(x+\sqrt{x^{2}+1}\right)
  2. Differentiate u. For the root use the power rule for functions: \frac{d}{dx}(x^{2}+1)^{1/2}=\frac{1}{2}(x^{2}+1)^{-1/2}(2x)=\frac{x}{\sqrt{x^{2}+1}}.
    \dfrac{d}{dx}\left(x+\sqrt{x^{2}+1}\right)=1+\dfrac{x}{\sqrt{x^{2}+1}}=\dfrac{\sqrt{x^{2}+1}+x}{\sqrt{x^{2}+1}}
  3. Substitute. The factor x+\sqrt{x^{2}+1} cancels.
    \dfrac{dy}{dx}=\dfrac{1}{x+\sqrt{x^{2}+1}}\cdot\dfrac{x+\sqrt{x^{2}+1}}{\sqrt{x^{2}+1}}=\dfrac{1}{\sqrt{x^{2}+1}}
\dfrac{dy}{dx}=\dfrac{1}{\sqrt{x^{2}+1}}
22
Find \frac{dy}{dx}: y=e^{-3x}\cos x
Solution
  1. Product rule, with \frac{d}{dx}e^{-3x}=-3e^{-3x} and \frac{d}{dx}\cos x=-\sin x.
    \dfrac{dy}{dx}=e^{-3x}(-\sin x)+\cos x\,(-3e^{-3x})
  2. Take out the common factor -e^{-3x}.
    \dfrac{dy}{dx}=-e^{-3x}(\sin x+3\cos x)
\dfrac{dy}{dx}=-e^{-3x}(\sin x+3\cos x)
23
Find \frac{dy}{dx} of the parametric functions: x=t+\dfrac{1}{t},\ y=t+1
Solution
  1. For parametric equations the chain rule gives \frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}, that is, divide \frac{dy}{dt} by \frac{dx}{dt} (as in Examples 35 and 36).
    \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}
  2. Differentiate x w.r.t. t.
    \dfrac{dx}{dt}=1-\dfrac{1}{t^{2}}=\dfrac{t^{2}-1}{t^{2}}
  3. Differentiate y w.r.t. t.
    \dfrac{dy}{dt}=1
  4. Divide.
    \dfrac{dy}{dx}=\dfrac{1}{(t^{2}-1)/t^{2}}=\dfrac{t^{2}}{t^{2}-1}
\dfrac{dy}{dx}=\dfrac{t^{2}}{t^{2}-1}
24
Find \frac{dy}{dx} of the parametric functions: x=t^{2}+\dfrac{1}{t^{2}},\ y=t-\dfrac{1}{t}
Solution
  1. For parametric equations the chain rule gives \frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}, that is, divide \frac{dy}{dt} by \frac{dx}{dt} (as in Examples 35 and 36).
    \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}
  2. Differentiate x w.r.t. t.
    \dfrac{dx}{dt}=2t-\dfrac{2}{t^{3}}=\dfrac{2(t^{4}-1)}{t^{3}}
  3. Differentiate y w.r.t. t.
    \dfrac{dy}{dt}=1+\dfrac{1}{t^{2}}=\dfrac{t^{2}+1}{t^{2}}
  4. Divide, then factor t^{4}-1=(t^{2}-1)(t^{2}+1) and cancel t^{2}+1.
    \dfrac{dy}{dx}=\dfrac{t^{2}+1}{t^{2}}\cdot\dfrac{t^{3}}{2(t^{4}-1)}=\dfrac{t(t^{2}+1)}{2(t^{2}-1)(t^{2}+1)}=\dfrac{t}{2(t^{2}-1)}
\dfrac{dy}{dx}=\dfrac{t}{2(t^{2}-1)}
25
Find \frac{dy}{dx} of the parametric functions: x=\dfrac{\theta^{2}-1}{\theta^{2}+1},\ y=\dfrac{\theta-1}{\theta+1}
Solution
  1. Here the parameter is \theta, so \frac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}.
  2. Quotient rule for x.
    \dfrac{dx}{d\theta}=\dfrac{(\theta^{2}+1)(2\theta)-(\theta^{2}-1)(2\theta)}{(\theta^{2}+1)^{2}}=\dfrac{4\theta}{(\theta^{2}+1)^{2}}
  3. Quotient rule for y.
    \dfrac{dy}{d\theta}=\dfrac{(\theta+1)(1)-(\theta-1)(1)}{(\theta+1)^{2}}=\dfrac{2}{(\theta+1)^{2}}
  4. Divide \frac{dy}{d\theta} by \frac{dx}{d\theta} (multiply by the reciprocal).
    \dfrac{dy}{dx}=\dfrac{2}{(\theta+1)^{2}}\cdot\dfrac{(\theta^{2}+1)^{2}}{4\theta}=\dfrac{(\theta^{2}+1)^{2}}{2\theta(\theta+1)^{2}}
\dfrac{dy}{dx}=\dfrac{(\theta^{2}+1)^{2}}{2\theta(\theta+1)^{2}}
26
Find \frac{dy}{dx} of the parametric functions: x=\sin 2\theta,\ y=\cos 4\theta
Solution
  1. Differentiate both with the chain rule.
    \dfrac{dx}{d\theta}=2\cos 2\theta\qquad \dfrac{dy}{d\theta}=-4\sin 4\theta
  2. Divide.
    \dfrac{dy}{dx}=\dfrac{-4\sin 4\theta}{2\cos 2\theta}=-\dfrac{2\sin 4\theta}{\cos 2\theta}
  3. Simplify with the double-angle formula \sin 4\theta=2\sin 2\theta\cos 2\theta.
    \dfrac{dy}{dx}=-\dfrac{2(2\sin 2\theta\cos 2\theta)}{\cos 2\theta}=-4\sin 2\theta
\dfrac{dy}{dx}=-4\sin 2\theta
27
Find \Delta y and dy: y=x^{2}+1
Solution
  1. By definition \Delta y=f(x+\Delta x)-f(x).
    \Delta y=\left[(x+\Delta x)^{2}+1\right]-\left[x^{2}+1\right]
  2. Expand (x+\Delta x)^{2}=x^{2}+2x\,\Delta x+(\Delta x)^{2}; the terms x^{2} and 1 cancel.
    \Delta y=2x\,\Delta x+(\Delta x)^{2}
  3. The differential is dy=f'(x)\,dx, and here f'(x)=2x.
    dy=2x\,dx
  4. Since dx=\Delta x, the two differ only by (\Delta x)^{2}, which is tiny when \Delta x is small. So \Delta y\approx dy.
    \Delta y-dy=(\Delta x)^{2}
With x=1 and an exaggerated \Delta x=1: the curve rises \Delta y=3 (from P to Q), the tangent rises only dy=2. The gap is (\Delta x)^{2}=1.
\Delta y=2x\,\Delta x+(\Delta x)^{2},\qquad dy=2x\,dx
28
Find \Delta y and dy: y=\sin x
Solution
  1. By definition:
    \Delta y=\sin(x+\Delta x)-\sin x
  2. Expand with \sin(A+B)=\sin A\cos B+\cos A\sin B and group the terms.
    \Delta y=\sin x\cos\Delta x+\cos x\sin\Delta x-\sin x=\cos x\sin\Delta x-\sin x\,(1-\cos\Delta x)
  3. The differential uses \frac{d}{dx}\sin x=\cos x.
    dy=\cos x\,dx
  4. For a small \Delta x (in radians), \sin\Delta x\approx\Delta x and \cos\Delta x\approx1, so \Delta y\approx\cos x\,\Delta x=dy.
\Delta y=\cos x\sin\Delta x-\sin x\,(1-\cos\Delta x),\qquad dy=\cos x\,dx
29
Use the concept of the differential to find the approximated value of the given expression: (1.8)^{5}
Solution
  1. Use the approximation of Section 2.16: for a small change \Delta x=dx,
    f(x+\Delta x)\approx f(x)+dy=f(x)+f'(x)\,dx
  2. Take f(x)=x^{5} and the nearby number x=2, whose fifth power is easy. Then dx=1.8-2=-0.2.
  3. Find f(2) and f'(2).
    f(2)=2^{5}=32,\qquad f'(x)=5x^{4}\Rightarrow f'(2)=5(16)=80
  4. Find the differential.
    dy=f'(2)\,dx=80(-0.2)=-16
  5. Add.
    (1.8)^{5}\approx 32+(-16)=16
(1.8)^{5}\approx 16
30
Use the concept of the differential to find the approximated value of the given expression: \sqrt{37}
Solution
  1. Use the approximation of Section 2.16: for a small change \Delta x=dx,
    f(x+\Delta x)\approx f(x)+dy=f(x)+f'(x)\,dx
  2. Take f(x)=\sqrt{x} with x=36 (a perfect square near 37) and dx=1.
  3. Find f(36) and f'(36).
    f(36)=6,\qquad f'(x)=\dfrac{1}{2\sqrt{x}}\Rightarrow f'(36)=\dfrac{1}{12}
  4. Find the differential.
    dy=\dfrac{1}{12}(1)=0.0833
  5. Add.
    \sqrt{37}\approx 6+0.0833=6.0833
\sqrt{37}\approx 6.083
31
Use the concept of the differential to find the approximated value of the given expression: \sin 31^\circ
Solution
  1. Use the approximation of Section 2.16: for a small change \Delta x=dx,
    f(x+\Delta x)\approx f(x)+dy=f(x)+f'(x)\,dx
  2. Take f(x)=\sin x with x=30^\circ=\frac{\pi}{6} and dx=1^\circ. The derivative formula \frac{d}{dx}\sin x=\cos x holds only in radians, so convert dx.
    dx=1^\circ=\dfrac{\pi}{180}=0.01745\ \text{rad}
  3. Find f and f' at 30^\circ.
    f(30^\circ)=\sin 30^\circ=0.5,\qquad f'(30^\circ)=\cos 30^\circ=\dfrac{\sqrt3}{2}=0.8660
  4. Find the differential.
    dy=0.8660\times0.01745=0.01511
  5. Add.
    \sin 31^\circ\approx 0.5+0.01511=0.5151
\sin 31^\circ\approx 0.5151
32
Use the concept of the differential to find the approximated value of the given expression: \tan\left(\dfrac{\pi}{4}+0.1\right)
Solution
  1. Use the approximation of Section 2.16: for a small change \Delta x=dx,
    f(x+\Delta x)\approx f(x)+dy=f(x)+f'(x)\,dx
  2. Take f(x)=\tan x with x=\frac{\pi}{4} and dx=0.1 (already in radians).
  3. Find f and f' at \frac{\pi}{4}.
    f\left(\dfrac{\pi}{4}\right)=\tan\dfrac{\pi}{4}=1,\qquad f'(x)=\sec^{2}x\Rightarrow f'\left(\dfrac{\pi}{4}\right)=(\sqrt2)^{2}=2
  4. Find the differential.
    dy=2(0.1)=0.2
  5. Add.
    \tan\left(\dfrac{\pi}{4}+0.1\right)\approx 1+0.2=1.2
\tan\left(\dfrac{\pi}{4}+0.1\right)\approx 1.2

Exercise 2.8p. 74

Practises higher order derivatives (Sections 2.17 and 2.18): differentiate the first derivative again to get \frac{d^{2}y}{dx^{2}}, and keep going for the third, fourth and higher derivatives.

1
Find the second derivative of the function w.r.t. the variable involved: y=-x^{3}+6x+9
Solution
  1. Differentiate once with the power rule; the constant 9 gives 0.
    \dfrac{dy}{dx}=-3x^{2}+6
  2. Differentiate the first derivative again.
    \dfrac{d^{2}y}{dx^{2}}=-6x
\dfrac{d^{2}y}{dx^{2}}=-6x
2
Find the second derivative of the function w.r.t. the variable involved: f(x)=30x^{2}-x^{3}
Solution
  1. First derivative (power rule).
    f'(x)=60x-3x^{2}
  2. Second derivative.
    f''(x)=60-6x
f''(x)=60-6x
3
Find the second derivative of the function w.r.t. the variable involved: f(x)=(-5x+9)^{2}
Solution
  1. Power rule for functions; the inside derivative is -5.
    f'(x)=2(-5x+9)(-5)=-10(-5x+9)=50x-90
  2. Differentiate again.
    f''(x)=50
f''(x)=50
4
Find the second derivative of the function w.r.t. the variable involved: y=2x^{6}+5x^{3}-6x^{2}
Solution
  1. Differentiate term by term.
    \dfrac{dy}{dx}=12x^{5}+15x^{2}-12x
  2. Differentiate again.
    \dfrac{d^{2}y}{dx^{2}}=60x^{4}+30x-12
\dfrac{d^{2}y}{dx^{2}}=60x^{4}+30x-12
5
Find the second derivative of the function w.r.t. the variable involved: y=20x^{-3}
Solution
  1. Power rule \frac{d}{dx}x^{n}=nx^{n-1} with n=-3.
    \dfrac{dy}{dx}=20(-3)x^{-4}=-60x^{-4}
  2. Differentiate again with n=-4.
    \dfrac{d^{2}y}{dx^{2}}=-60(-4)x^{-5}=240x^{-5}=\dfrac{240}{x^{5}}
\dfrac{d^{2}y}{dx^{2}}=\dfrac{240}{x^{5}}

Note. The book prints y=20x-3 with the exponent lowered. It is read as y=20x^{-3}; if it really were 20x-3, then \frac{dy}{dx}=20 and \frac{d^{2}y}{dx^{2}}=0.

6
Find the second derivative of the function w.r.t. the variable involved: y=\dfrac{2}{x^{4}}
Solution
  1. Write y=2x^{-4} and use the power rule.
    \dfrac{dy}{dx}=-8x^{-5}
  2. Differentiate again.
    \dfrac{d^{2}y}{dx^{2}}=40x^{-6}=\dfrac{40}{x^{6}}
\dfrac{d^{2}y}{dx^{2}}=\dfrac{40}{x^{6}}
7
Find the second derivative of the function w.r.t. the variable involved: f(x)=x^{2}(3x-4)^{3}
Solution
  1. Product rule, with \frac{d}{dx}(3x-4)^{3}=3(3x-4)^{2}(3)=9(3x-4)^{2}.
    f'(x)=2x(3x-4)^{3}+9x^{2}(3x-4)^{2}
  2. Take out the common factor x(3x-4)^{2}; inside, 2(3x-4)+9x=15x-8.
    f'(x)=x(3x-4)^{2}\left[2(3x-4)+9x\right]=(3x-4)^{2}(15x^{2}-8x)
  3. Product rule again on (3x-4)^{2}\cdot(15x^{2}-8x), with \frac{d}{dx}(3x-4)^{2}=6(3x-4) and \frac{d}{dx}(15x^{2}-8x)=30x-8.
    f''(x)=6(3x-4)(15x^{2}-8x)+(3x-4)^{2}(30x-8)
  4. Take out (3x-4). The bracket is 6(15x^{2}-8x)+(3x-4)(30x-8)=90x^{2}-48x+90x^{2}-144x+32.
    f''(x)=(3x-4)(180x^{2}-192x+32)=4(3x-4)(45x^{2}-48x+8)
f''(x)=4(3x-4)(45x^{2}-48x+8)
8
Find the second derivative of the function w.r.t. the variable involved: f(x)=(x^{2}+5x-1)^{4}
Solution
  1. Power rule for functions.
    f'(x)=4(x^{2}+5x-1)^{3}(2x+5)
  2. Product rule on (x^{2}+5x-1)^{3}\cdot(2x+5); the first factor gives 3(x^{2}+5x-1)^{2}(2x+5) and the second gives 2.
    f''(x)=4\left[3(x^{2}+5x-1)^{2}(2x+5)^{2}+2(x^{2}+5x-1)^{3}\right]
  3. Take out (x^{2}+5x-1)^{2}. The bracket is 3(4x^{2}+20x+25)+2(x^{2}+5x-1)=14x^{2}+70x+73.
    f''(x)=4(x^{2}+5x-1)^{2}(14x^{2}+70x+73)
f''(x)=4(x^{2}+5x-1)^{2}(14x^{2}+70x+73)
9
Find the second derivative of the function w.r.t. the variable involved: f(x)=\cos 10x
Solution
  1. Chain rule: \frac{d}{dx}\cos 10x=-\sin 10x\cdot 10.
    f'(x)=-10\sin 10x
  2. Chain rule again: \frac{d}{dx}\sin 10x=10\cos 10x.
    f''(x)=-100\cos 10x
f''(x)=-100\cos 10x
10
Find the second derivative of the function w.r.t. the variable involved: f(x)=\tan\dfrac{x}{2}
Solution
  1. Chain rule: \frac{d}{dx}\tan u=\sec^{2}u\,\frac{du}{dx} with u=\frac{x}{2}, \frac{du}{dx}=\frac{1}{2}.
    f'(x)=\dfrac{1}{2}\sec^{2}\dfrac{x}{2}
  2. \sec^{2}\frac{x}{2}=\left(\sec\frac{x}{2}\right)^{2}: power rule for functions, then \frac{d}{dx}\sec u=\sec u\tan u\,\frac{du}{dx}.
    f''(x)=\dfrac{1}{2}\cdot 2\sec\dfrac{x}{2}\cdot\sec\dfrac{x}{2}\tan\dfrac{x}{2}\cdot\dfrac{1}{2}=\dfrac{1}{2}\sec^{2}\dfrac{x}{2}\tan\dfrac{x}{2}
f''(x)=\dfrac{1}{2}\sec^{2}\dfrac{x}{2}\tan\dfrac{x}{2}
11
Find the second derivative of the function w.r.t. the variable involved: f(\theta)=\sin^{2}5\theta
Solution
  1. Power rule for functions, then the chain rule for \sin 5\theta. Finally use 2\sin A\cos A=\sin 2A to simplify.
    f'(\theta)=2\sin 5\theta\cdot 5\cos 5\theta=10\sin 5\theta\cos 5\theta=5\sin 10\theta
  2. Differentiate 5\sin 10\theta.
    f''(\theta)=50\cos 10\theta
f''(\theta)=50\cos 10\theta
12
Find the second derivative of the function w.r.t. the variable involved: f(\theta)=\dfrac{1}{3+2\cos\theta}
Solution
  1. Write f(\theta)=(3+2\cos\theta)^{-1} and use the power rule for functions; the inside derivative is -2\sin\theta.
    f'(\theta)=-(3+2\cos\theta)^{-2}(-2\sin\theta)=\dfrac{2\sin\theta}{(3+2\cos\theta)^{2}}
  2. Quotient rule on f'(\theta): top 2\sin\theta (derivative 2\cos\theta), bottom (3+2\cos\theta)^{2} (derivative 2(3+2\cos\theta)(-2\sin\theta)).
    f''(\theta)=\dfrac{(3+2\cos\theta)^{2}(2\cos\theta)-2\sin\theta\cdot 2(3+2\cos\theta)(-2\sin\theta)}{(3+2\cos\theta)^{4}}
  3. Cancel one factor (3+2\cos\theta) from the top and bottom.
    f''(\theta)=\dfrac{2\cos\theta(3+2\cos\theta)+8\sin^{2}\theta}{(3+2\cos\theta)^{3}}=\dfrac{6\cos\theta+4\cos^{2}\theta+8\sin^{2}\theta}{(3+2\cos\theta)^{3}}
  4. Use \sin^{2}\theta=1-\cos^{2}\theta so the top has \cos\theta only.
    f''(\theta)=\dfrac{8+6\cos\theta-4\cos^{2}\theta}{(3+2\cos\theta)^{3}}=\dfrac{2(4+3\cos\theta-2\cos^{2}\theta)}{(3+2\cos\theta)^{3}}
f''(\theta)=\dfrac{2(4+3\cos\theta-2\cos^{2}\theta)}{(3+2\cos\theta)^{3}}
13
Find the second derivative of the function w.r.t. the variable involved: f(x)=e^{2x}(x^{2}+1)
Solution
  1. Product rule, with \frac{d}{dx}e^{2x}=2e^{2x}. Then take out 2e^{2x}.
    f'(x)=2e^{2x}(x^{2}+1)+e^{2x}(2x)=2e^{2x}(x^{2}+x+1)
  2. Product rule again on 2e^{2x}(x^{2}+x+1), then collect terms.
    f''(x)=2\left[2e^{2x}(x^{2}+x+1)+e^{2x}(2x+1)\right]=2e^{2x}(2x^{2}+4x+3)
f''(x)=2e^{2x}(2x^{2}+4x+3)
14
Find the second derivative of the function w.r.t. the variable involved: f(x)=(x^{2}+1)\ln(x^{2}+1)
Solution
  1. Product rule, with \frac{d}{dx}\ln(x^{2}+1)=\frac{2x}{x^{2}+1}.
    f'(x)=2x\ln(x^{2}+1)+(x^{2}+1)\cdot\dfrac{2x}{x^{2}+1}=2x\ln(x^{2}+1)+2x
  2. Differentiate again; use the product rule on 2x\ln(x^{2}+1).
    f''(x)=2\ln(x^{2}+1)+2x\cdot\dfrac{2x}{x^{2}+1}+2=2\ln(x^{2}+1)+\dfrac{4x^{2}}{x^{2}+1}+2
f''(x)=2\ln(x^{2}+1)+\dfrac{4x^{2}}{x^{2}+1}+2
15
Find the indicated derivative: y=4x^{7}+x^{6}-x^{4}; \dfrac{d^{4}y}{dx^{4}}
Solution
  1. Differentiate term by term with the power rule.
    \dfrac{dy}{dx}=28x^{6}+6x^{5}-4x^{3}
  2. Second derivative.
    \dfrac{d^{2}y}{dx^{2}}=168x^{5}+30x^{4}-12x^{2}
  3. Third derivative.
    \dfrac{d^{3}y}{dx^{3}}=840x^{4}+120x^{3}-24x
  4. Fourth derivative.
    \dfrac{d^{4}y}{dx^{4}}=3360x^{3}+360x^{2}-24
\dfrac{d^{4}y}{dx^{4}}=3360x^{3}+360x^{2}-24
16
Find the indicated derivative: y=\dfrac{2}{x}; \dfrac{d^{5}y}{dx^{5}}
Solution
  1. Write y=2x^{-1} so that the power rule can be used again and again (as in Example 42).
  2. First derivative.
    \dfrac{dy}{dx}=-2x^{-2}
  3. Second derivative.
    \dfrac{d^{2}y}{dx^{2}}=4x^{-3}
  4. Third derivative.
    \dfrac{d^{3}y}{dx^{3}}=-12x^{-4}
  5. Fourth derivative.
    \dfrac{d^{4}y}{dx^{4}}=48x^{-5}
  6. Fifth derivative.
    \dfrac{d^{5}y}{dx^{5}}=-240x^{-6}=-\dfrac{240}{x^{6}}
\dfrac{d^{5}y}{dx^{5}}=-\dfrac{240}{x^{6}}
17
Find the indicated derivative: f(x)=\cos\pi x; f'''(x)
Solution
  1. Each differentiation brings out a factor \pi (chain rule, since \frac{d}{dx}(\pi x)=\pi).
  2. First derivative.
    f'(x)=-\pi\sin\pi x
  3. Second derivative.
    f''(x)=-\pi^{2}\cos\pi x
  4. Third derivative.
    f'''(x)=\pi^{3}\sin\pi x
f'''(x)=\pi^{3}\sin\pi x
18
Find the indicated derivative: f(x)=\dfrac{1}{\sec(2x+1)}; f^{(4)}(x)
Solution
  1. Simplify first: \frac{1}{\sec\theta}=\cos\theta.
    f(x)=\cos(2x+1)
  2. First derivative (chain rule, inside derivative 2).
    f'(x)=-2\sin(2x+1)
  3. Second derivative.
    f''(x)=-4\cos(2x+1)
  4. Third derivative.
    f'''(x)=8\sin(2x+1)
  5. Fourth derivative.
    f^{(4)}(x)=16\cos(2x+1)
f^{(4)}(x)=16\cos(2x+1)
19
Let f(x)=x^{3}+2x.
Solution
(a) Find f'(x) and f''(x).
  1. Power rule, term by term.
    f'(x)=3x^{2}+2
  2. Differentiate again.
    f''(x)=6x
f'(x)=3x^{2}+2,\qquad f''(x)=6x
(b) In general, f''(x)=\lim\limits_{\Delta x\to0}\dfrac{f'(x+\Delta x)-f'(x)}{\Delta x}, provided the limit exists. Use f'(x) obtained in part (a) and use the definition to find f''(x).
  1. The second derivative is the derivative of f', so apply the definition of the derivative to f'(x)=3x^{2}+2.
    f''(x)=\lim_{\Delta x\to0}\dfrac{f'(x+\Delta x)-f'(x)}{\Delta x}
  2. Substitute f'(x+\Delta x)=3(x+\Delta x)^{2}+2.
    f''(x)=\lim_{\Delta x\to0}\dfrac{\left[3(x+\Delta x)^{2}+2\right]-\left[3x^{2}+2\right]}{\Delta x}
  3. Expand (x+\Delta x)^{2}=x^{2}+2x\,\Delta x+(\Delta x)^{2}; the terms 3x^{2} and 2 cancel.
    f''(x)=\lim_{\Delta x\to0}\dfrac{6x\,\Delta x+3(\Delta x)^{2}}{\Delta x}
  4. Divide by \Delta x (allowed because \Delta x\neq0 before the limit), then let \Delta x\to0.
    f''(x)=\lim_{\Delta x\to0}(6x+3\Delta x)=6x
  5. This agrees with the answer of part (a).
f''(x)=6x
20
Show that \dfrac{d^{2}}{dx^{2}}(fg)=f''g+2f'g'+fg'' and \dfrac{d^{3}}{dx^{3}}(fg)=f'''g+3f''g'+3f'g''+fg'''.
Solution

Here f and g are functions of x that can be differentiated three times. The only tool needed is the product rule (uv)'=u'v+uv', used again and again.

Second derivative
  1. Product rule once.
    \dfrac{d}{dx}(fg)=f'g+fg'
  2. Differentiate again. Each term is itself a product, so use the product rule on each.
    \dfrac{d^{2}}{dx^{2}}(fg)=\left(f''g+f'g'\right)+\left(f'g'+fg''\right)
  3. Collect the two equal middle terms.
    \dfrac{d^{2}}{dx^{2}}(fg)=f''g+2f'g'+fg''
Third derivative
  1. Differentiate the result for the second derivative, term by term.
    \dfrac{d^{3}}{dx^{3}}(fg)=\dfrac{d}{dx}(f''g)+2\dfrac{d}{dx}(f'g')+\dfrac{d}{dx}(fg'')
  2. Use the product rule on each of the three terms.
    =\left(f'''g+f''g'\right)+2\left(f''g'+f'g''\right)+\left(f'g''+fg'''\right)
  3. Collect like terms: f''g' appears 1+2=3 times and f'g'' appears 2+1=3 times.
    \dfrac{d^{3}}{dx^{3}}(fg)=f'''g+3f''g'+3f'g''+fg'''
(fg)''=f''g+2f'g'+fg'',\qquad (fg)'''=f'''g+3f''g'+3f'g''+fg'''

Exercise 2.9p. 80–81

Practises extrema of functions (Sections 2.19 to 2.21): critical values, absolute extrema on a closed interval, concavity, points of inflection and the second derivative test.

1
Find the critical values of the function.
Solution
(i) f(x)=2x^{2}-6x+8
  1. Differentiate.
    f'(x)=4x-6
  2. f'(x) is a polynomial, so it exists for every x. Set it equal to zero.
    4x-6=0\Rightarrow x=\dfrac{3}{2}
  3. x=\frac32 is in the domain \mathbb{R}, so it is a critical value. The point on the graph is \left(\frac32,\frac72\right), the lowest point of the parabola.
The tangent is horizontal at the critical value.
x=\dfrac{3}{2}
(ii) f(x)=x^{3}+x-2
  1. Differentiate.
    f'(x)=3x^{2}+1
  2. f'(x) exists for every x. Setting 3x^{2}+1=0 gives x^{2}=-\frac13, which has no real solution because a square cannot be negative.
    3x^{2}+1\ge1>0\ \text{ for all }x
  3. So there is no number where f'(x)=0 or where f'(x) fails to exist: f has no critical values. It is increasing everywhere.
The graph rises all the time and never has a horizontal tangent.
\text{no critical values}
(iii) f(x)=\dfrac{x}{x^{2}+2}
  1. Quotient rule.
    f'(x)=\dfrac{(x^{2}+2)(1)-x(2x)}{(x^{2}+2)^{2}}=\dfrac{2-x^{2}}{(x^{2}+2)^{2}}
  2. The denominator (x^{2}+2)^{2} is never zero, so f'(x) exists for all x. Set the numerator equal to zero.
    2-x^{2}=0\Rightarrow x=\pm\sqrt2
  3. Both values are in the domain, so both are critical values. They give the turning points \left(\sqrt2,\frac{\sqrt2}{4}\right) and \left(-\sqrt2,-\frac{\sqrt2}{4}\right).
x=-\sqrt2,\quad x=\sqrt2
(iv) f(x)=\cos 4x
  1. Chain rule.
    f'(x)=-4\sin 4x
  2. f'(x) exists for every x. Set it equal to zero; \sin\theta=0 exactly when \theta is a whole multiple of \pi.
    -4\sin 4x=0\Rightarrow 4x=n\pi\Rightarrow x=\dfrac{n\pi}{4},\quad n\in\mathbb{Z}
  3. So the critical values are \dots,-\frac{\pi}{4},0,\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\pi,\dots At each of them the graph reaches its highest value 1 or its lowest value -1.
Critical values are spaced \frac{\pi}{4} apart, at every peak and every trough.
x=\dfrac{n\pi}{4},\quad n=0,\pm1,\pm2,\dots
(v) f(x)=(4x-3)^{1/3}
  1. Power rule for functions.
    f'(x)=\dfrac{1}{3}(4x-3)^{-2/3}(4)=\dfrac{4}{3(4x-3)^{2/3}}
  2. The numerator 4 is never zero, so f'(x)\neq0 anywhere. But f'(x) does not exist where the denominator is zero.
    4x-3=0\Rightarrow x=\dfrac{3}{4}
  3. x=\frac34 is in the domain of f (the cube root of 0 is 0, so f(\frac34)=0). Hence x=\frac34 is a critical value. There the graph has a vertical tangent.
At x=\frac34 the tangent is vertical, so f' does not exist there.
x=\dfrac{3}{4}
(vi) f(x)=x^{2}(x+1)^{3}
  1. Product rule.
    f'(x)=2x(x+1)^{3}+3x^{2}(x+1)^{2}
  2. Take out the common factor x(x+1)^{2}; inside, 2(x+1)+3x=5x+2.
    f'(x)=x(x+1)^{2}(5x+2)
  3. f'(x) is a polynomial, so it always exists. Set each factor equal to zero.
    x=0,\qquad x+1=0\Rightarrow x=-1,\qquad 5x+2=0\Rightarrow x=-\dfrac{2}{5}
  4. All three are in the domain, so the critical values are -1, -\frac25 and 0.
Horizontal tangents at all three critical values; at x=-1 the curve flattens but keeps rising.
x=-1,\quad x=-\dfrac{2}{5},\quad x=0
2
Find the absolute extrema of the function on the indicated interval.
Solution
(i) f(x)=-x^{2}+6x;\ [1,4]
  1. Differentiate.
    f'(x)=-2x+6
  2. Critical value: -2x+6=0 gives x=3, which lies in [1,4].
  3. Evaluate f at the critical value and at both end points.
    f(1)=-1+6=5,\qquad f(3)=-9+18=9,\qquad f(4)=-16+24=8
  4. Compare the three values: the largest is 9 and the smallest is 5.
\text{absolute max}=f(3)=9,\qquad \text{absolute min}=f(1)=5
(ii) f(x)=(x-1)^{2};\ [2,5]
  1. Differentiate.
    f'(x)=2(x-1)
  2. f'(x)=0 only at x=1, which is not in [2,5]. So only the end points need to be checked.
  3. Evaluate at the end points.
    f(2)=(2-1)^{2}=1,\qquad f(5)=(5-1)^{2}=16
  4. On [2,5], f'(x)>0, so f is increasing: the smallest value is at the left end and the largest at the right end.
\text{absolute max}=f(5)=16,\qquad \text{absolute min}=f(2)=1
(iii) f(x)=x^{2/3};\ [-1,8]
  1. Differentiate with the power rule.
    f'(x)=\dfrac{2}{3}x^{-1/3}=\dfrac{2}{3\sqrt[3]{x}}
  2. f'(x) is never zero, but it does not exist at x=0, which lies in [-1,8] and where f(0)=0 is defined. So x=0 is the critical value.
  3. Evaluate at the critical value and the end points. Note (-1)^{2/3}=\left(\sqrt[3]{-1}\right)^{2}=(-1)^{2}=1.
    f(-1)=1,\qquad f(0)=0,\qquad f(8)=\left(\sqrt[3]{8}\right)^{2}=4
  4. Compare the three values: the largest is 4 and the smallest is 0.
The minimum is at the sharp point (cusp) x=0, where f' does not exist.
\text{absolute max}=f(8)=4,\qquad \text{absolute min}=f(0)=0
(iv) f(x)=x^{3}-6x^{2}+2;\ [-3,2]
  1. Differentiate and factor.
    f'(x)=3x^{2}-12x=3x(x-4)
  2. f'(x)=0 gives x=0 or x=4. Only x=0 lies in [-3,2].
  3. Evaluate at x=0 and at both end points.
    f(-3)=-27-54+2=-79,\qquad f(0)=2,\qquad f(2)=8-24+2=-14
  4. Compare: the largest value is 2 and the smallest is -79.
\text{absolute max}=f(0)=2,\qquad \text{absolute min}=f(-3)=-79
(v) f(x)=1+5\sin 3x;\ \left[0,\tfrac{\pi}{2}\right]
  1. Differentiate (chain rule).
    f'(x)=15\cos 3x
  2. Set f'(x)=0. For 0\le x\le\frac{\pi}{2} the angle 3x runs from 0 to \frac{3\pi}{2}, and \cos 3x=0 there only when 3x=\frac{\pi}{2} or 3x=\frac{3\pi}{2}.
    x=\dfrac{\pi}{6}\quad\text{or}\quad x=\dfrac{\pi}{2}
  3. Evaluate at the critical values and at the end points (x=\frac{\pi}{2} is both).
    \begin{aligned}f(0)&=1+5\sin 0=1\\ f\left(\dfrac{\pi}{6}\right)&=1+5\sin\dfrac{\pi}{2}=6\\ f\left(\dfrac{\pi}{2}\right)&=1+5\sin\dfrac{3\pi}{2}=-4\end{aligned}
  4. Compare: the largest value is 6 and the smallest is -4.
\text{absolute max}=f\left(\dfrac{\pi}{6}\right)=6,\qquad \text{absolute min}=f\left(\dfrac{\pi}{2}\right)=-4
(vi) f(x)=2\cos 2x-4\cos x;\ [0,2\pi]
  1. Differentiate.
    f'(x)=-4\sin 2x+4\sin x
  2. Use \sin 2x=2\sin x\cos x and factor.
    f'(x)=-8\sin x\cos x+4\sin x=4\sin x(1-2\cos x)
  3. Set each factor equal to zero in [0,2\pi].
    \sin x=0\Rightarrow x=0,\ \pi,\ 2\pi;\qquad \cos x=\dfrac{1}{2}\Rightarrow x=\dfrac{\pi}{3},\ \dfrac{5\pi}{3}
  4. Evaluate f at all these points (the end points 0 and 2\pi are already in the list).
    \begin{aligned}f(0)&=2-4=-2\\ f\left(\dfrac{\pi}{3}\right)&=2\left(-\dfrac12\right)-4\left(\dfrac12\right)=-3\\ f(\pi)&=2(1)-4(-1)=6\end{aligned}
  5. In the same way f\left(\frac{5\pi}{3}\right)=2\left(-\frac12\right)-4\left(\frac12\right)=-3 and f(2\pi)=-2.
  6. Compare: the largest value is 6 and the smallest is -3 (reached at two points).
\text{absolute max}=f(\pi)=6,\qquad \text{absolute min}=f\left(\dfrac{\pi}{3}\right)=f\left(\dfrac{5\pi}{3}\right)=-3
3
Use the second derivative to determine the intervals on which the function is concave upward and concave downward.
Solution
(i) f(x)=-x^{2}+7x
  1. First derivative.
    f'(x)=-2x+7
  2. Second derivative.
    f''(x)=-2
  3. f''(x)=-2<0 for every x, so the graph is concave downward on the whole real line and never concave upward.
A parabola opening downward bends down everywhere.
\text{concave downward on }(-\infty,\infty);\ \text{never concave upward}
(ii) f(x)=-x^{3}+6x^{2}+x-1
  1. First derivative.
    f'(x)=-3x^{2}+12x+1
  2. Second derivative.
    f''(x)=-6x+12=-6(x-2)
  3. f''(x)=0 at x=2. Test the sign on each side.
    x<2:\ f''(x)>0;\qquad x>2:\ f''(x)<0
  4. So the graph is concave upward to the left of 2 and concave downward to the right. The concavity changes at (2,f(2))=(2,17).
\text{concave upward on }(-\infty,2),\qquad \text{concave downward on }(2,\infty)
(iii) f(x)=(x+5)^{5}
  1. First derivative (power rule for functions).
    f'(x)=5(x+5)^{4}
  2. Second derivative.
    f''(x)=20(x+5)^{3}
  3. (x+5)^{3} has the same sign as x+5.
    x<-5:\ f''(x)<0;\qquad x>-5:\ f''(x)>0
The curve bends down to the left of x=-5 and up to the right.
\text{concave downward on }(-\infty,-5),\qquad \text{concave upward on }(-5,\infty)
(iv) f(x)=x(x-4)^{3}
  1. First derivative: product rule, then factor.
    f'(x)=(x-4)^{3}+3x(x-4)^{2}=(x-4)^{2}(4x-4)
  2. Second derivative: product rule on (x-4)^{2}(4x-4), then factor.
    f''(x)=2(x-4)(4x-4)+4(x-4)^{2}=(x-4)(12x-24)=12(x-4)(x-2)
  3. f''(x)=0 at x=2 and x=4. A sign table shows the concavity on each interval.
Intervalx-2x-4f''(x)Concavity
(-\infty,2)--+upward
(2,4)+--downward
(4,\infty)+++upward
Concavity changes at x=2 and x=4 (dashed lines).
\text{concave upward on }(-\infty,2)\text{ and }(4,\infty),\qquad \text{concave downward on }(2,4)
(v) f(x)=x^{1/2}+2x
  1. The domain is x\ge0 because of the square root.
  2. First derivative.
    f'(x)=\dfrac{1}{2}x^{-1/2}+2
  3. Second derivative.
    f''(x)=-\dfrac{1}{4}x^{-3/2}=-\dfrac{1}{4x\sqrt{x}}
  4. For every x>0 the denominator 4x\sqrt x is positive, so f''(x)<0.
The curve lies above the chord joining its ends, so it bends downward.
\text{concave downward on }(0,\infty);\ \text{never concave upward}
(vi) f(x)=x+\dfrac{9}{x}
  1. First derivative.
    f'(x)=1-\dfrac{9}{x^{2}}
  2. Second derivative.
    f''(x)=\dfrac{18}{x^{3}}
  3. x^{3} has the same sign as x, and x=0 is not in the domain.
    x<0:\ f''(x)<0;\qquad x>0:\ f''(x)>0
  4. There is no point of inflection at x=0, because f is not defined there.
\text{concave downward on }(-\infty,0),\qquad \text{concave upward on }(0,\infty)
4
Use the second derivative to locate all points of inflection.
Solution
(i) f(x)=x^{4}-x^{3}+2x^{2}+x-1
  1. First derivative.
    f'(x)=4x^{3}-3x^{2}+4x+1
  2. Second derivative.
    f''(x)=12x^{2}-6x+4
  3. Try to solve f''(x)=0. The discriminant is negative, so there is no real root.
    b^{2}-4ac=(-6)^{2}-4(12)(4)=36-192=-156<0
  4. A quadratic with no real roots keeps one sign, and its leading coefficient 12>0, so f''(x)>0 for all x. The graph is concave upward everywhere and has no point of inflection.
The curve bends upward everywhere.
\text{no point of inflection}
(ii) f(x)=x^{5/3}+4x
  1. First derivative.
    f'(x)=\dfrac{5}{3}x^{2/3}+4
  2. Second derivative.
    f''(x)=\dfrac{10}{9}x^{-1/3}=\dfrac{10}{9\sqrt[3]{x}}
  3. f''(x) is never zero, but it does not exist at x=0, where f is continuous (f(0)=0). Check the sign on each side.
    x<0:\ f''(x)<0\ (\text{concave down});\qquad x>0:\ f''(x)>0\ (\text{concave up})
  4. The concavity changes at x=0, so (0,0) is a point of inflection.
The curve crosses its tangent y=4x at the point of inflection.
(0,\,0)
(iii) f(x)=\sin x
  1. Differentiate twice.
    f'(x)=\cos x,\qquad f''(x)=-\sin x
  2. f''(x)=0 when \sin x=0, that is x=n\pi (n an integer).
  3. \sin x changes sign at every multiple of \pi, so f'' changes sign there and the concavity changes. At these points f(n\pi)=\sin n\pi=0.
Every point where the curve crosses the x-axis is a point of inflection.
(n\pi,\,0),\quad n=0,\pm1,\pm2,\dots
(iv) f(x)=\cos x
  1. Differentiate twice.
    f'(x)=-\sin x,\qquad f''(x)=-\cos x
  2. f''(x)=0 when \cos x=0, that is x=\frac{\pi}{2}+n\pi.
  3. \cos x changes sign at each of these points, so the concavity changes. There f(x)=\cos x=0.
Again the points of inflection are where the curve crosses the x-axis.
\left(\dfrac{\pi}{2}+n\pi,\,0\right),\quad n=0,\pm1,\pm2,\dots
(v) f(x)=x-\sin x
  1. Differentiate twice.
    f'(x)=1-\cos x,\qquad f''(x)=\sin x
  2. f''(x)=0 when x=n\pi, and \sin x changes sign at each multiple of \pi, so each gives a point of inflection.
  3. The y-coordinate is f(n\pi)=n\pi-\sin n\pi=n\pi.
The points of inflection (n\pi,n\pi) lie on the line y=x.
(n\pi,\,n\pi),\quad n=0,\pm1,\pm2,\dots
(vi) f(x)=\tan x
  1. Differentiate twice. For \sec^{2}x=(\sec x)^{2} use the power rule for functions.
    f'(x)=\sec^{2}x,\qquad f''(x)=2\sec x\cdot\sec x\tan x=2\sec^{2}x\tan x
  2. \sec^{2}x>0 wherever \tan x is defined, so f'' has the same sign as \tan x. So f''(x)=0 when \tan x=0, that is x=n\pi.
  3. \tan x changes from negative to positive at each x=n\pi, so the concavity changes there, and f(n\pi)=0. At x=\frac{\pi}{2}+n\pi the function is not defined (vertical asymptotes), so no inflection point occurs there.
(n\pi,\,0),\quad n=0,\pm1,\pm2,\dots
5
Use second derivative test to find the relative extrema of the function.
Solution
(i) f(x)=-(-2x-5)^{2}
  1. Differentiate (power rule for functions; the inside derivative is -2).
    f'(x)=-2(-2x-5)(-2)=4(-2x-5)=-8x-20
  2. Critical value: -8x-20=0 gives x=-\frac{5}{2}.
  3. Second derivative.
    f''(x)=-8
  4. f''\left(-\frac52\right)=-8<0, so f has a relative maximum at x=-\frac52.
    f\left(-\dfrac{5}{2}\right)=-\left(5-5\right)^{2}=0
\text{relative maximum }f\left(-\dfrac{5}{2}\right)=0
(ii) f(x)=x^{3}+3x^{2}+3x+1
  1. First derivative.
    f'(x)=3x^{2}+6x+3=3(x+1)^{2}
  2. Critical value: 3(x+1)^{2}=0 gives x=-1.
  3. Second derivative.
    f''(x)=6x+6
  4. f''(-1)=0, so the second derivative test fails (it gives no decision).
  5. Use the sign of f' instead: f'(x)=3(x+1)^{2}\ge0 on both sides of -1, so f keeps increasing and there is no relative extremum. In fact f(x)=(x+1)^{3}, and (-1,0) is a point of inflection.
The graph flattens at (-1,0) but does not turn, so there is no maximum or minimum.
\text{no relative extrema}
(iii) f(x)=6x^{5}-10x^{2}
  1. First derivative.
    f'(x)=30x^{4}-20x=10x(3x^{3}-2)
  2. Critical values: x=0, or 3x^{3}=2, which gives x=\sqrt[3]{\frac{2}{3}}\approx0.874.
  3. Second derivative.
    f''(x)=120x^{3}-20
  4. Test x=0: f''(0)=-20<0, so there is a relative maximum f(0)=0.
  5. Test x=\sqrt[3]{2/3}: here x^{3}=\frac23, so f''=120\cdot\frac23-20=60>0, a relative minimum. To find its value use x^{5}=x^{3}\cdot x^{2}=\frac23x^{2}.
    f=6\cdot\dfrac{2}{3}x^{2}-10x^{2}=-6x^{2}=-6\left(\dfrac{2}{3}\right)^{2/3}\approx-4.58
\text{rel. max }f(0)=0,\qquad \text{rel. min }f\left(\sqrt[3]{\tfrac{2}{3}}\right)=-6\left(\tfrac{2}{3}\right)^{2/3}\approx-4.58
(iv) f(x)=x^{2}+\dfrac{1}{x^{2}}
  1. First derivative.
    f'(x)=2x-\dfrac{2}{x^{3}}=\dfrac{2(x^{4}-1)}{x^{3}}
  2. Critical values: x^{4}=1 gives x=\pm1. (x=0 is not in the domain, so it is not a critical value.)
  3. Second derivative.
    f''(x)=2+\dfrac{6}{x^{4}}
  4. f''(\pm1)=2+6=8>0, so both are relative minima.
    f(1)=f(-1)=1+1=2
\text{relative minimum }f(-1)=f(1)=2
(v) f(x)=\cos 3x,\ [0,2\pi]
  1. First derivative.
    f'(x)=-3\sin 3x
  2. f'(x)=0 when \sin 3x=0, so 3x=n\pi. Inside (0,2\pi) this gives:
    x=\dfrac{\pi}{3},\ \dfrac{2\pi}{3},\ \pi,\ \dfrac{4\pi}{3},\ \dfrac{5\pi}{3}
  3. Second derivative.
    f''(x)=-9\cos 3x
  4. Test each critical value (table below). At the end points x=0 and x=2\pi, f=1 as well, but the test is applied only at critical values inside the interval.
x\cos 3xf''(x)Conclusionf(x)
\frac{\pi}{3}-19>0rel. min-1
\frac{2\pi}{3}1-9<0rel. max1
\pi-19>0rel. min-1
\frac{4\pi}{3}1-9<0rel. max1
\frac{5\pi}{3}-19>0rel. min-1
Maxima 1 at \frac{2\pi}{3},\frac{4\pi}{3}; minima -1 at \frac{\pi}{3},\pi,\frac{5\pi}{3}.
\text{rel. max}=1\text{ at }x=\tfrac{2\pi}{3},\tfrac{4\pi}{3};\qquad \text{rel. min}=-1\text{ at }x=\tfrac{\pi}{3},\pi,\tfrac{5\pi}{3}
(vi) f(x)=\cos x+\sin x,\ [0,2\pi]
  1. First derivative.
    f'(x)=-\sin x+\cos x
  2. f'(x)=0 gives \sin x=\cos x, that is \tan x=1. In [0,2\pi]:
    x=\dfrac{\pi}{4}\quad\text{or}\quad x=\dfrac{5\pi}{4}
  3. Second derivative.
    f''(x)=-\cos x-\sin x
  4. At x=\frac{\pi}{4}: f''=-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}=-\sqrt2<0, so there is a relative maximum f\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\sqrt2.
  5. At x=\frac{5\pi}{4}: f''=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\sqrt2>0, so there is a relative minimum f\left(\frac{5\pi}{4}\right)=-\sqrt2.
\text{rel. max }f\left(\dfrac{\pi}{4}\right)=\sqrt2,\qquad \text{rel. min }f\left(\dfrac{5\pi}{4}\right)=-\sqrt2
6
Determine whether the given function has a relative extremum at the indicated points.
Solution
(i) f(x)=\cos x\sin x,\ x=\dfrac{\pi}{4}
  1. Write f(x)=\frac12\sin 2x (double-angle formula) and differentiate.
    f'(x)=\cos 2x
  2. f'\left(\frac{\pi}{4}\right)=\cos\frac{\pi}{2}=0, so \frac{\pi}{4} is a critical value.
  3. Second derivative.
    f''(x)=-2\sin 2x
  4. f''\left(\frac{\pi}{4}\right)=-2\sin\frac{\pi}{2}=-2<0, so f has a relative maximum at x=\frac{\pi}{4}, with value f\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}\cdot\frac{\sqrt2}{2}=\frac12.
\text{relative maximum at }x=\dfrac{\pi}{4};\ f\left(\dfrac{\pi}{4}\right)=\dfrac{1}{2}
(ii) f(x)=x\sin x,\ x=0
  1. Product rule.
    f'(x)=\sin x+x\cos x
  2. f'(0)=0+0=0, so x=0 is a critical value.
  3. Differentiate again.
    f''(x)=\cos x+\cos x-x\sin x=2\cos x-x\sin x
  4. f''(0)=2>0, so f has a relative minimum at x=0, with value f(0)=0.
\text{relative minimum at }x=0;\ f(0)=0
(iii) f(x)=\tan^{2}x,\ x=\pi
  1. Power rule for functions.
    f'(x)=2\tan x\sec^{2}x
  2. f'(\pi)=2(0)(1)=0, so x=\pi is a critical value.
  3. Product rule on 2\tan x\cdot\sec^{2}x, using \frac{d}{dx}\sec^{2}x=2\sec^{2}x\tan x.
    f''(x)=2\sec^{2}x\cdot\sec^{2}x+2\tan x\cdot 2\sec^{2}x\tan x=2\sec^{4}x+4\sec^{2}x\tan^{2}x
  4. At x=\pi: \sec^{2}\pi=1 and \tan\pi=0, so f''(\pi)=2>0. Hence f has a relative minimum at x=\pi, with f(\pi)=0.
\text{relative minimum at }x=\pi;\ f(\pi)=0
(iv) f(x)=(1+\sin x)^{3},\ x=\dfrac{\pi}{8}
  1. Power rule for functions.
    f'(x)=3(1+\sin x)^{2}\cos x
  2. At x=\frac{\pi}{8} both factors are positive: 1+\sin\frac{\pi}{8}\approx1.383 and \cos\frac{\pi}{8}\approx0.924.
    f'\left(\dfrac{\pi}{8}\right)=3(1.383)^{2}(0.924)\approx5.30\neq0
  3. A relative extremum of a differentiable function can occur only where f'=0. Here f'\left(\frac{\pi}{8}\right)>0, so f is increasing at x=\frac{\pi}{8} and there is no relative extremum there. (The critical values of f are where \cos x=0, such as \pm\frac{\pi}{2}.)
At x=\frac{\pi}{8} the tangent (dashed) slopes upward, so the point is not a maximum or a minimum.
\text{no relative extremum at }x=\dfrac{\pi}{8},\ \text{since }f'\left(\tfrac{\pi}{8}\right)\approx5.30\neq0

Exercise 2.10p. 85–87

Applications of derivatives (Section 2.22): limits and continuity in real problems, differentials for errors, related rates, optimisation, inflation (Price Growth Model P=P_{0}e^{rt}) and straight-line models.

1
According to Einstein's theory of relativity, the mass m of a body moving with velocity v is m=\dfrac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}, where m_{0} is the initial mass and c is the speed of light. What happens to m as v\to c^{-}?
Solution
  1. As v increases towards c from below, \frac{v^{2}}{c^{2}} increases towards 1, so 1-\frac{v^{2}}{c^{2}} decreases towards 0 through positive values.
    v\to c^{-}\ \Rightarrow\ \dfrac{v^{2}}{c^{2}}\to1^{-}\ \Rightarrow\ 1-\dfrac{v^{2}}{c^{2}}\to0^{+}
  2. The square root of a small positive number is a small positive number. So the denominator tends to 0^{+} while the numerator stays the fixed positive number m_{0}.
    \sqrt{1-\dfrac{v^{2}}{c^{2}}}\to0^{+}
  3. A fixed positive number divided by a positive number that tends to zero grows without bound.
    \lim_{v\to c^{-}}m=\lim_{v\to c^{-}}\dfrac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=+\infty
  4. Only the left-hand limit makes sense: for v>c the number under the root is negative, so m would not be real.
The mass grows faster and faster as the speed approaches the speed of light.
v/c0.50.90.990.999
m/m_{0}1.1552.2947.08922.37
The line v=c is a vertical asymptote: m\to\infty as v\to c^{-}.
\lim_{v\to c^{-}}m=+\infty
2
The function below is continuous at x=3. What is k?
f(x)=\begin{cases}kx+1, & x\le 3\\ 2-kx, & x>3\end{cases}
Solution
  1. For continuity at x=3: f(3) must be defined, \lim_{x\to3}f(x) must exist, and the two must be equal. So the left-hand limit, the right-hand limit and f(3) must be the same number.
  2. Value at 3 (use the top rule, since x\le3).
    f(3)=3k+1
  3. Left-hand limit (values x<3 use kx+1).
    \lim_{x\to3^{-}}f(x)=\lim_{x\to3^{-}}(kx+1)=3k+1
  4. Right-hand limit (values x>3 use 2-kx).
    \lim_{x\to3^{+}}f(x)=\lim_{x\to3^{+}}(2-kx)=2-3k
  5. Equate the two one-sided limits and solve.
    3k+1=2-3k\Rightarrow 6k=1\Rightarrow k=\dfrac{1}{6}
  6. Check: with k=\frac16 both limits and f(3) equal 3\cdot\frac16+1=\frac32.
With k=\frac16 the two pieces meet at (3,\frac32), so there is no jump.
k=\dfrac{1}{6}
3
The volume v of a sphere of radius r is v=\left(\dfrac{4\pi}{3}\right)r^{3}. Find the surface area s of the sphere if s is the instantaneous rate of change of the volume with respect to the radius.
Solution
  1. The instantaneous rate of change of v with respect to r is the derivative \frac{dv}{dr}.
  2. Differentiate with the power rule; \frac{4\pi}{3} is a constant.
    s=\dfrac{dv}{dr}=\dfrac{4\pi}{3}\cdot3r^{2}=4\pi r^{2}
  3. This is the familiar formula for the surface area of a sphere.
s=4\pi r^{2}
4
The height S above ground of a projectile at time t is given by S(t)=\dfrac{1}{2}gt^{2}+v_{0}t+s_{0}, where g, v_{0} and s_{0} are constants. Find the instantaneous rate of change of S with respect to t at t=4.
Solution
  1. The instantaneous rate of change of S with respect to t is S'(t). Differentiate term by term; g, v_{0} and s_{0} are constants, so \frac{d}{dt}s_{0}=0.
    S'(t)=\dfrac{d}{dt}\left(\dfrac{1}{2}gt^{2}+v_{0}t+s_{0}\right)=gt+v_{0}
  2. Put t=4.
    S'(4)=4g+v_{0}
  3. S'(t) is the velocity of the projectile, so this is its velocity at t=4.
S'(4)=4g+v_{0}
5
The side of a square is measured to be 10\,\mathrm{cm} with a possible error of \pm0.3\,\mathrm{cm}. Use differentials to find an approximation to the maximum error in the area. Find the approximate relative error and the approximate percentage error.
Solution
Given
  • x = 10\,\mathrm{cm} measured side
  • dx = \pm0.3\,\mathrm{cm} possible error in the side
To find
  • maximum error dA
  • relative error
  • percentage error
  1. Area of a square of side x.
    A=x^{2}
  2. Differential of the area (Section 2.15): dA=A'(x)\,dx.
    dA=2x\,dx
  3. Substitute x=10 and dx=\pm0.3.
    dA=2(10)(\pm0.3)=\pm6\,\mathrm{cm^{2}}
  4. Relative error = error \div actual area. The area is A=10^{2}=100\,\mathrm{cm^{2}}.
    \dfrac{dA}{A}=\dfrac{\pm6}{100}=\pm0.06
  5. Percentage error = relative error \times100\%.
    \pm0.06\times100\%=\pm6\%
dA=\pm6\,\mathrm{cm^{2}},\qquad \dfrac{dA}{A}=\pm0.06,\qquad \text{percentage error}=\pm6\%
6
A woman jogging at a constant rate of 10\,\mathrm{km/hr} crosses a point A heading north. Ten minutes later a man jogging at a constant rate of 9\,\mathrm{km/hr} crosses the same point heading east. How fast is the distance between the joggers changing 20 minutes after the man crosses A?
The joggers and the right triangle they form with A.
Solution
  1. Let x be the man's distance east of A, y the woman's distance north of A, and z the distance between them. North and east are at right angles, so by Pythagoras:
    z^{2}=x^{2}+y^{2}
  2. Differentiate w.r.t. time t, since every length changes with time.
    2z\dfrac{dz}{dt}=2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}\ \Rightarrow\ \dfrac{dz}{dt}=\dfrac{x\dfrac{dx}{dt}+y\dfrac{dy}{dt}}{z}
  3. Positions at the required moment. The man has run for 20 minutes, the woman for 10+20=30 minutes.
    x=9\times\dfrac{20}{60}=3\,\mathrm{km},\qquad y=10\times\dfrac{30}{60}=5\,\mathrm{km}
  4. Distance between them.
    z=\sqrt{3^{2}+5^{2}}=\sqrt{34}\,\mathrm{km}
  5. Substitute \frac{dx}{dt}=9 and \frac{dy}{dt}=10.
    \dfrac{dz}{dt}=\dfrac{3(9)+5(10)}{\sqrt{34}}=\dfrac{77}{\sqrt{34}}\approx13.2\,\mathrm{km/hr}
  6. The rate is positive, so the joggers are moving apart.
\dfrac{dz}{dt}=\dfrac{77}{\sqrt{34}}\approx13.2\,\mathrm{km/hr}

Note. The book prints 'How fast is the difference between the joggers hanging'; it means 'how fast is the distance between the joggers changing'.

7
A plate in the shape of an equilateral triangle is expanding with time. A side increases at a constant rate of 2\,\mathrm{cm/hr}. At what rate is the area increasing when the side is 8\,\mathrm{cm}?
Solution
An equilateral triangle of side x has height \frac{\sqrt3}{2}x.
  1. Area = \frac12\times base \times height, with base x and height \frac{\sqrt3}{2}x.
    A=\dfrac{1}{2}\cdot x\cdot\dfrac{\sqrt3}{2}x=\dfrac{\sqrt3}{4}x^{2}
  2. Differentiate w.r.t. time (chain rule).
    \dfrac{dA}{dt}=\dfrac{\sqrt3}{4}\cdot2x\dfrac{dx}{dt}=\dfrac{\sqrt3}{2}x\dfrac{dx}{dt}
  3. Substitute x=8 and \frac{dx}{dt}=2.
    \dfrac{dA}{dt}=\dfrac{\sqrt3}{2}(8)(2)=8\sqrt3\approx13.86\,\mathrm{cm^{2}/hr}
\dfrac{dA}{dt}=8\sqrt3\approx13.9\,\mathrm{cm^{2}/hr}
8
A rectangle expands with time. The diagonal of the rectangle increases at a rate of 1\,\mathrm{in/hr} and the length increases at a rate of \frac{1}{4}\,\mathrm{in/hr}. How fast is its width increasing when the width is 6\,\mathrm{in} and the length is 8\,\mathrm{in}?
Solution
Length, width and diagonal form a right triangle.
  1. Let the length be x, the width y and the diagonal D. By Pythagoras:
    x^{2}+y^{2}=D^{2}
  2. Differentiate w.r.t. t and divide by 2.
    x\dfrac{dx}{dt}+y\dfrac{dy}{dt}=D\dfrac{dD}{dt}
  3. At this moment x=8 and y=6, so D=\sqrt{64+36}=10. Also \frac{dx}{dt}=\frac14 and \frac{dD}{dt}=1.
    8\left(\dfrac{1}{4}\right)+6\dfrac{dy}{dt}=10(1)
  4. Solve.
    6\dfrac{dy}{dt}=10-2=8\Rightarrow\dfrac{dy}{dt}=\dfrac{4}{3}\approx1.33\,\mathrm{in/hr}
\dfrac{dy}{dt}=\dfrac{4}{3}\approx1.33\,\mathrm{in/hr}

Note. The book prints the rate of the length as '\frac14 min/hr'; it should be \frac14 in/hr.

9
The side of a cube increases at a rate of 5\,\mathrm{cm/hr}. At what rate does the diagonal of the cube increase?
Solution
The main diagonal D is the hypotenuse of a right triangle with legs x\sqrt2 (face diagonal) and x (edge).
  1. Let the side be x. The diagonal of the bottom face is \sqrt{x^{2}+x^{2}}=x\sqrt2. The main diagonal D joins opposite corners of the cube; Pythagoras in the right triangle formed by the face diagonal, a vertical edge and D gives
    D^{2}=(x\sqrt2)^{2}+x^{2}=3x^{2}\Rightarrow D=\sqrt3\,x
  2. Differentiate w.r.t. t.
    \dfrac{dD}{dt}=\sqrt3\,\dfrac{dx}{dt}
  3. Substitute \frac{dx}{dt}=5.
    \dfrac{dD}{dt}=5\sqrt3\approx8.66\,\mathrm{cm/hr}
  4. The rate is the same at every moment, because D is a fixed multiple of x.
\dfrac{dD}{dt}=5\sqrt3\approx8.66\,\mathrm{cm/hr}
10
A particle moves on the graph of y^{2}=x+1 so that \dfrac{dx}{dt}=4x+4. What is \dfrac{dx}{dt} when x=8?
Solution
  1. As printed, the answer comes straight from the given rule.
    \dfrac{dx}{dt}=4(8)+4=36
  2. The intended question (the standard version of this problem) asks for \frac{dy}{dt} when x=8. Differentiate the equation of the curve w.r.t. t.
    2y\dfrac{dy}{dt}=\dfrac{dx}{dt}
  3. When x=8: y^{2}=9, so y=3 or y=-3, and \frac{dx}{dt}=36.
    \dfrac{dy}{dt}=\dfrac{36}{2y}=\dfrac{36}{\pm6}=\pm6
  4. So at (8,3) the particle rises with \frac{dy}{dt}=6, and at (8,-3) it falls with \frac{dy}{dt}=-6.
Both points of the curve with x=8.
\dfrac{dx}{dt}=36;\qquad \dfrac{dy}{dt}=6\text{ at }(8,3),\quad \dfrac{dy}{dt}=-6\text{ at }(8,-3)

Note. The book asks for \frac{dx}{dt} at x=8, which the given rule gives at once (36). It is probably a misprint for \frac{dy}{dt}, so both are answered.

11
At 8:00 am ship S_{1} is 20\,\mathrm{km} due north of S_{2}. Ship S_{1} sails south at a rate of 9\,\mathrm{km/hr} and S_{2} sails west at a rate of 12\,\mathrm{km/hr}. At 9:20 am, at what rate is the distance between the two ships changing?
Solution
Positions at 9:20 am: S_{1} is y=8\,\mathrm{km} north of O and S_{2} is x=16\,\mathrm{km} west of O.
  1. Take O as the position of S_{2} at 8:00 am. After t hours, S_{2} is x km west of O and S_{1} is y km north of O.
    x=12t,\qquad y=20-9t
  2. West and north are at right angles, so the distance D between the ships satisfies
    D^{2}=x^{2}+y^{2}
  3. Differentiate w.r.t. t and divide by 2. Here \frac{dx}{dt}=12 and \frac{dy}{dt}=-9 (y decreases because S_{1} sails south).
    D\dfrac{dD}{dt}=x\dfrac{dx}{dt}+y\dfrac{dy}{dt}
  4. At 9:20 am, t=1\frac13=\frac43 hours.
    \begin{aligned}x&=12\cdot\dfrac{4}{3}=16\\ y&=20-9\cdot\dfrac{4}{3}=8\\ D&=\sqrt{16^{2}+8^{2}}=\sqrt{320}=8\sqrt5\end{aligned}
  5. Substitute.
    \dfrac{dD}{dt}=\dfrac{16(12)+8(-9)}{8\sqrt5}=\dfrac{120}{8\sqrt5}=\dfrac{15}{\sqrt5}=3\sqrt5\approx6.71\,\mathrm{km/hr}
  6. The rate is positive, so the distance between the ships is increasing at about 6.71\,\mathrm{km/hr}.
\dfrac{dD}{dt}=3\sqrt5\approx6.71\,\mathrm{km/hr}\ (\text{increasing})
12
Find two non-negative numbers whose sum is 60 and whose product is a maximum.
Solution
  1. Let the numbers be x and y with x+y=60, x\ge0 and y\ge0. Then y=60-x and 0\le x\le60.
  2. Write the product as a function of x only.
    P(x)=x(60-x)=60x-x^{2}
  3. Critical value.
    P'(x)=60-2x=0\Rightarrow x=30
  4. Second derivative test: P''(x)=-2<0, so x=30 gives a maximum. (At the end points P(0)=P(60)=0.)
  5. Then y=60-30=30 and the maximum product is 30\times30=900.
The product is largest when the two numbers are equal.
30\text{ and }30;\quad \text{maximum product}=900
13
If the total fence to be used is 8000\,\mathrm{m}, find the dimensions of the enclosed land in the figure that has the greatest area.
The land is fenced on all sides and divided by two fences parallel to the width.
Solution
  1. From the figure, let x be the length and y the width. The fence consists of 2 lengths and 4 widths (the two ends and the two dividers).
    2x+4y=8000\Rightarrow x=4000-2y
  2. Area to be maximised.
    A=xy=(4000-2y)y=4000y-2y^{2},\qquad 0<y<2000
  3. Critical value.
    \dfrac{dA}{dy}=4000-4y=0\Rightarrow y=1000
  4. Second derivative: \frac{d^{2}A}{dy^{2}}=-4<0, so this is a maximum.
  5. Find the length and the area.
    x=4000-2(1000)=2000,\qquad A=2000\times1000=2\times10^{6}\,\mathrm{m^{2}}
  6. Only the number of dividers matters, not where they stand.
Area A (in millions of \mathrm{m^{2}}) against the width y (in m): the peak is at y=1000\,\mathrm{m}.
\text{length }2000\,\mathrm{m},\ \text{width }1000\,\mathrm{m}\ (\text{area }2\times10^{6}\,\mathrm{m^{2}})
14
An open rectangular box is to be constructed with a square base and a volume of 32\,000\,\mathrm{cm^{3}}. Find the dimensions of the box that require the least amount of material.
Solution
Open box: square base of side x, height h, no top.
  1. Let the base be x by x and the height h (in cm). The fixed volume gives h in terms of x.
    x^{2}h=32000\Rightarrow h=\dfrac{32000}{x^{2}}
  2. The box has no top, so the material is the base plus four sides.
    S=x^{2}+4xh=x^{2}+4x\cdot\dfrac{32000}{x^{2}}=x^{2}+\dfrac{128000}{x}
  3. Critical value.
    S'(x)=2x-\dfrac{128000}{x^{2}}=0\Rightarrow x^{3}=64000\Rightarrow x=40
  4. Second derivative test.
    S''(x)=2+\dfrac{256000}{x^{3}}>0\ \text{ for }x>0
  5. So x=40 gives the least material.
    h=\dfrac{32000}{40^{2}}=\dfrac{32000}{1600}=20,\qquad S=1600+\dfrac{128000}{40}=4800\,\mathrm{cm^{2}}
\text{base }40\,\mathrm{cm}\times40\,\mathrm{cm},\ \text{height }20\,\mathrm{cm}
15
A company determines that for the production of x units of a commodity its revenue and cost functions are, respectively, R(x)=-3x^{2}+970x and C(x)=2x^{2}+500. Find the maximum profit and minimum average cost.
Solution
Maximum profit
  1. Profit = revenue - cost.
    P(x)=R(x)-C(x)=-3x^{2}+970x-2x^{2}-500=-5x^{2}+970x-500
  2. Critical value.
    P'(x)=-10x+970=0\Rightarrow x=97
  3. P''(x)=-10<0, so x=97 gives the maximum profit.
    P(97)=-5(97)^{2}+970(97)-500=-47045+94090-500=46545
Minimum average cost
  1. Average cost per unit = total cost \div number of units.
    \bar C(x)=\dfrac{C(x)}{x}=2x+\dfrac{500}{x}
  2. Critical value.
    \bar C'(x)=2-\dfrac{500}{x^{2}}=0\Rightarrow x^{2}=250\Rightarrow x=5\sqrt{10}\approx15.8
  3. \bar C''(x)=\frac{1000}{x^{3}}>0 for x>0, so this is a minimum.
    \bar C_{\min}=2(5\sqrt{10})+\dfrac{500}{5\sqrt{10}}=10\sqrt{10}+10\sqrt{10}=20\sqrt{10}\approx63.25
  4. If x must be a whole number of units, compare x=15 and x=16: \bar C(15)\approx63.33 and \bar C(16)=63.25. So 16 units give the least average cost, about 63.25 per unit.
P_{\max}=46545\ \text{at }x=97,\qquad \bar C_{\min}=20\sqrt{10}\approx63.25\ \text{at }x=5\sqrt{10}\approx15.8

Note. The book prints 'its remove and cost functions' and G(x); these are the revenue R(x) and the cost C(x).

16
If the inflation rate is continuously compounded at 4\% per year and the price of a commodity is $50 today:
Solution
(a) Derive the function for the price of the commodity over time.
  1. Continuous inflation at rate r means the price grows at a rate proportional to the price itself.
    \dfrac{dP}{dt}=rP,\qquad r=4\%=0.04
  2. The function whose derivative is 0.04 times itself is the exponential of the Price Growth Model, P(t)=P_{0}e^{rt}. Here P_{0}=50.
    P(t)=50e^{0.04t}
  3. Check: P'(t)=50(0.04)e^{0.04t}=0.04P(t) and P(0)=50e^{0}=50.
P(t)=50e^{0.04t}
(b) Find the price after 8 years.
  1. Put t=8.
    P(8)=50e^{0.04(8)}=50e^{0.32}=50(1.3771)\approx68.86
P(8)\approx\$68.86
(c) Find the instantaneous rate of change of the price at t=8 years.
  1. Differentiate the price function.
    P'(t)=50(0.04)e^{0.04t}=2e^{0.04t}
  2. Put t=8.
    P'(8)=2e^{0.32}=2(1.3771)\approx2.75
  3. After 8 years the price is rising at about $2.75 per year.
The slope of the tangent at t=8 is the rate of change of price.
P'(8)\approx\$2.75\text{ per year}
17
A company models its operational cost as C(t)=500e^{0.04t}-100t, where t is the time in years.
Solution
(a) Find the rate of change of cost at any time t.
  1. Differentiate; \frac{d}{dt}e^{0.04t}=0.04e^{0.04t}.
    C'(t)=500(0.04)e^{0.04t}-100=20e^{0.04t}-100
C'(t)=20e^{0.04t}-100
(b) Determine when the rate of increase in cost is minimal.
  1. The cost is least when it stops falling and starts rising, that is, when its rate of increase C'(t) is zero.
    20e^{0.04t}-100=0\Rightarrow e^{0.04t}=5\Rightarrow t=\dfrac{\ln5}{0.04}=25\ln5\approx40.2\text{ years}
  2. Second derivative: C''(t)=0.8e^{0.04t}>0, so this is a minimum of the cost. Before t\approx40.2 the cost falls (C'<0); after it, the cost rises (C'>0).
  3. Because C''(t)>0 for every t, the rate C'(t) itself keeps increasing. So the rate of change is smallest at the start: C'(0)=20-100=-80 per year.
t=25\ln5\approx40.2\text{ years}

Note. The book's wording 'Determining the rate of increase in cost is minimal' is unclear. It is answered as 'when is the cost minimal' (C'(t)=0, t\approx40.2 years); the smallest rate of change, -80 per year at t=0, is also shown. The model gives a negative cost at t\approx40.2 (C\approx-1524), so it is only realistic for the early years.

18
The price of a commodity P(t) is given by P(t)=150(1+0.05t)^{2}, where t is measured in years and P(t) is the price level.
Solution
(a) Find the instantaneous rate of change of prices at t=3 years.
  1. Power rule for functions; the inside derivative is 0.05.
    P'(t)=150\cdot2(1+0.05t)(0.05)=15(1+0.05t)
  2. Put t=3.
    P'(3)=15(1.15)=17.25
P'(3)=17.25\text{ per year}
(b) Calculate the inflation rate at t=3 years.
  1. The inflation rate is the rate of change of price as a fraction of the price (in the Price Growth Model \frac{dP}{dt}=rP, so r=\frac{P'(t)}{P(t)}).
    r=\dfrac{P'(3)}{P(3)}\times100\%
  2. Price at t=3.
    P(3)=150(1.15)^{2}=150(1.3225)=198.375
  3. Inflation rate.
    r=\dfrac{17.25}{198.375}\times100\%\approx8.70\%\text{ per year}
r\approx8.70\%\text{ per year}
19
A ship sails in a straight line. Its distance d(t) (in nautical miles) from port is modelled by d(t)=15t, where t is time in hours.
Solution
(a) Find the speed of the ship.
  1. Speed is the rate of change of distance, d'(t).
    d'(t)=\dfrac{d}{dt}(15t)=15
\text{speed}=15\text{ nautical miles per hour}
(b) Calculate the distance after 3 hours.
  1. Put t=3.
    d(3)=15(3)=45
d(3)=45\text{ nautical miles}
(c) Explain the meaning of the slope.

The graph of d(t)=15t is a straight line through the origin with slope 15. The slope is the rate of change of distance with time, so it is the speed: the ship moves 15 nautical miles farther from port every hour. Because the slope is constant, the speed is constant (uniform motion).

Every hour the distance increases by 15 nautical miles: the slope is the speed.
\text{slope}=15=\text{constant speed}
20
A cyclist is travelling along a straight path, and the distance travelled s(t) (in metres) is given by s(t)=5t^{2}+3t, where t is the time in seconds.
Solution
(a) Find the speed at any time t.
  1. Speed is the derivative of distance with respect to time.
    v(t)=s'(t)=10t+3\,\mathrm{m\,s^{-1}}
v(t)=10t+3\,\mathrm{m\,s^{-1}}
(b) Determine the speed at t=4 seconds.
  1. Put t=4.
    v(4)=10(4)+3=43\,\mathrm{m\,s^{-1}}
v(4)=43\,\mathrm{m\,s^{-1}}
(c) Interpret the significance of the slope in this context.

The slope of the tangent to the distance-time graph at any instant is the speed at that instant. Here the graph is a parabola that gets steeper: the slope 10t+3 grows with time, so the cyclist is speeding up. The slope increases by 10\,\mathrm{m\,s^{-1}} every second, so the acceleration is s''(t)=10\,\mathrm{m\,s^{-2}}.

At t=4 the tangent has slope 43, the speed in \mathrm{m\,s^{-1}}.
\text{slope of tangent}=\text{speed};\ \text{increasing slope}\Rightarrow\text{accelerating}

Review exercisep. 87–88

Mixed questions on the whole unit: limits, continuity, derivatives, differentials, related rates, extrema and curve sketching.

1 (i)
If \lim\limits_{x\to a}f(x)=3 and \lim\limits_{x\to a}g(x)=0, then \lim\limits_{x\to a}\dfrac{f(x)}{g(x)}:
  1. A3
  2. B0
  3. CExist
  4. DDoesn't existAnswer
Why

The quotient theorem \lim\frac{f}{g}=\frac{\lim f}{\lim g} needs \lim g\neq0, so it cannot be used here. Near x=a the top is close to 3 while the bottom is close to 0, so \frac{f(x)}{g(x)} becomes larger and larger in size (for example \frac{3}{0.001}=3000). It does not approach any real number, so the limit does not exist.

Options (a) and (b) come from wrongly treating \frac{3}{0} as a number.

1 (ii)
If f(x) is defined as below, then \lim\limits_{x\to0^{-}}f(x) is:
f(x)=\begin{cases}2x-1, & x<0\\ 2x+1, & x>0\end{cases}
  1. A1
  2. B-1Answer
  3. C0
  4. D2
Why

For x\to0^{-} we use values x<0, so the rule is 2x-1: \lim\limits_{x\to0^{-}}(2x-1)=2(0)-1=-1. (The right-hand limit is +1, so \lim\limits_{x\to0}f(x) itself does not exist.)

Note. The book prints '\lim_{x\to0^{-}}f(x)=0, is'; the '=0' is a misprint.

1 (iii)
If f and g are continuous at 2, then \dfrac{f}{g} is continuous at:
  1. A0
  2. B1
  3. C2Answer
  4. D3
Why

A quotient of functions that are continuous at a number is continuous at that number, provided the denominator is not zero there. So \frac{f}{g} is continuous at 2 (assuming g(2)\neq0). Nothing is known about f and g at 0, 1 or 3.

1 (iv)
The function f below is continuous at:
f(x)=\begin{cases}\dfrac{\sin x}{x}, & x\neq0\\ 1, & x=0\end{cases}
  1. A0Answer
  2. B1
  3. C-1
  4. D0.1
Why

The only point where continuity must be tested is x=0, where the rule changes. There \lim\limits_{x\to0}\frac{\sin x}{x}=1=f(0), so f is continuous at 0.

(At 1, -1 and 0.1, f(x)=\frac{\sin x}{x} is also continuous, being a quotient of continuous functions with a non-zero denominator. The question is testing the special point x=0.)

1 (v)
If f is differentiable for every value of x, then f is:
  1. ADiscontinuous
  2. BContinuousAnswer
  3. CFinite
  4. DInfinite
Why

Section 2.8: if f'(a) exists then f is continuous at a. So a function differentiable everywhere is continuous everywhere. (The converse is false: |x| is continuous at 0 but not differentiable there.)

1 (vi)
If k is a constant and n is a positive integer, then \dfrac{d}{dx}k^{n} is:
  1. Ank^{n-1}
  2. Bk^{n-1}
  3. C\ln n\cdot k^{n}
  4. D0Answer
Why

k^{n} contains no x, so it is a constant, and the derivative of a constant is 0. Option (a) is a trap: the power rule \frac{d}{dx}x^{n}=nx^{n-1} is for the variable x, not for a constant.

1 (vii)
If f(2)=2 and g(x)=x^{2}, then \dfrac{d}{dx}\left[\dfrac{3g(x)}{f(2)}\right] is:
  1. A2x
  2. B3xAnswer
  3. C\frac{3}{2}x
  4. D\frac{3}{2}x^{2}
Why

f(2)=2 is just a number, so \frac{3g(x)}{f(2)}=\frac{3x^{2}}{2}. Its derivative is \frac{3}{2}\cdot2x=3x.

1 (viii)
If y=f(x) is a polynomial function of degree 2, then \dfrac{d^{3}}{dx^{3}}f(x) is:
  1. A0Answer
  2. B1
  3. C-1
  4. D2
Why

Let f(x)=ax^{2}+bx+c. Then f'(x)=2ax+b, f''(x)=2a (a constant) and f'''(x)=0. Each differentiation lowers the degree by one.

1 (ix)
If f is differentiable for every value of x, then f is continuous for:
  1. ASome value of x
  2. B[0,\infty]
  3. CEvery value of xAnswer
  4. D[0,-\infty]
Why

Differentiability at a point implies continuity at that point. Since f is differentiable at every x, it is continuous at every x.

1 (x)
If f(t)=2t^{3}, the absolute minimum is at:
  1. A3
  2. B0
  3. C-1Answer
  4. D1
Why

f'(t)=6t^{2}\ge0, so f is increasing. On a closed interval an increasing function has its smallest value at the left end point. Among the given values (the interval [-1,3]) the smallest is f(-1)=-2.

Option (b) is a trap: f'(0)=0, but f' does not change sign at 0, so t=0 is a point of inflection, not a minimum.

Note. The book prints 'If f(t)=2t^{3} is absolute minimum at'. On the whole real line 2t^{3} has no absolute minimum, so the question must mean the interval formed by the options, [-1,3].

2
Evaluate \lim\limits_{x\to0}\dfrac{x^{3}}{\sin^{2}3x}.
Solution
  1. Direct substitution gives \frac{0}{0}, so rearrange the expression to use the standard limit \lim\limits_{\theta\to0}\frac{\sin\theta}{\theta}=1 (equivalently \lim\limits_{\theta\to0}\frac{\theta}{\sin\theta}=1).
  2. Split x^{3}=x\cdot x^{2} and write \frac{x^{2}}{\sin^{2}3x}=\left(\frac{x}{\sin 3x}\right)^{2}. Then make the angle 3x appear on top.
    \dfrac{x^{3}}{\sin^{2}3x}=x\left(\dfrac{x}{\sin 3x}\right)^{2}=x\left(\dfrac{1}{3}\cdot\dfrac{3x}{\sin 3x}\right)^{2}=\dfrac{x}{9}\left(\dfrac{3x}{\sin 3x}\right)^{2}
  3. As x\to0, 3x\to0, so \frac{3x}{\sin 3x}\to1, while \frac{x}{9}\to0.
    \lim_{x\to0}\dfrac{x^{3}}{\sin^{2}3x}=\lim_{x\to0}\dfrac{x}{9}\cdot\left(\lim_{x\to0}\dfrac{3x}{\sin 3x}\right)^{2}=0\cdot1^{2}=0
\lim_{x\to0}\dfrac{x^{3}}{\sin^{2}3x}=0
3
Find the derivative of y=\dfrac{1+\sin x}{x\cos x}.
Solution
  1. Quotient rule with u=1+\sin x and v=x\cos x. Find u' and v' (product rule for v).
    u'=\cos x,\qquad v'=\cos x-x\sin x
  2. Substitute into \frac{dy}{dx}=\frac{vu'-uv'}{v^{2}}.
    \dfrac{dy}{dx}=\dfrac{x\cos x\cdot\cos x-(1+\sin x)(\cos x-x\sin x)}{x^{2}\cos^{2}x}
  3. Expand the numerator and use \sin^{2}x+\cos^{2}x=1 on the terms x\cos^{2}x+x\sin^{2}x.
    x\cos^{2}x-\cos x+x\sin x-\sin x\cos x+x\sin^{2}x=x+x\sin x-\cos x-\sin x\cos x
  4. Factor by grouping: x(1+\sin x)-\cos x(1+\sin x)=(1+\sin x)(x-\cos x).
    \dfrac{dy}{dx}=\dfrac{(1+\sin x)(x-\cos x)}{x^{2}\cos^{2}x}
\dfrac{dy}{dx}=\dfrac{(1+\sin x)(x-\cos x)}{x^{2}\cos^{2}x}
4
If f'(0)=-1 and g'(0)=6, what is \dfrac{d^{2}}{dx^{2}}\left[xf(x)+xg(x)\right] at x=0?
Solution
  1. Let h(x)=xf(x)+xg(x)=x\,F(x), where F(x)=f(x)+g(x).
  2. First derivative (product rule).
    h'(x)=F(x)+xF'(x)
  3. Second derivative (product rule on xF'(x)).
    h''(x)=F'(x)+F'(x)+xF''(x)=2F'(x)+xF''(x)
  4. At x=0 the last term is zero.
    h''(0)=2F'(0)=2\left[f'(0)+g'(0)\right]=2(-1+6)=10
\dfrac{d^{2}}{dx^{2}}\left[xf(x)+xg(x)\right]_{x=0}=10
5
Find \dfrac{d^{2}y}{dx^{2}} when x^{3}+y^{3}=27.
Solution
  1. Differentiate implicitly w.r.t. x.
    3x^{2}+3y^{2}\dfrac{dy}{dx}=0\Rightarrow\dfrac{dy}{dx}=-\dfrac{x^{2}}{y^{2}}
  2. Differentiate \frac{dy}{dx}=-\frac{x^{2}}{y^{2}} again with the quotient rule, remembering that y is a function of x.
    \dfrac{d^{2}y}{dx^{2}}=-\dfrac{y^{2}(2x)-x^{2}\left(2y\dfrac{dy}{dx}\right)}{y^{4}}
  3. Substitute \frac{dy}{dx}=-\frac{x^{2}}{y^{2}}, then multiply the top and bottom by y.
    \dfrac{d^{2}y}{dx^{2}}=-\dfrac{2xy^{2}+\dfrac{2x^{4}}{y}}{y^{4}}=-\dfrac{2xy^{3}+2x^{4}}{y^{5}}=-\dfrac{2x(y^{3}+x^{3})}{y^{5}}
  4. Use the equation of the curve, x^{3}+y^{3}=27.
    \dfrac{d^{2}y}{dx^{2}}=-\dfrac{2x(27)}{y^{5}}=-\dfrac{54x}{y^{5}}
\dfrac{d^{2}y}{dx^{2}}=-\dfrac{54x}{y^{5}}
6
Use differentials to find an approximate value of \sqrt{65}.
Solution
  1. Take f(x)=\sqrt{x} with x=64 (a perfect square near 65) and dx=1. Use f(x+dx)\approx f(x)+f'(x)\,dx.
  2. Find f(64) and f'(64).
    f(64)=8,\qquad f'(x)=\dfrac{1}{2\sqrt x}\Rightarrow f'(64)=\dfrac{1}{16}
  3. Differential.
    dy=\dfrac{1}{16}(1)=0.0625
  4. Add.
    \sqrt{65}\approx8+0.0625=8.0625
\sqrt{65}\approx8.0625
7
An oil storage tank in the form of a circular cylinder has a height of 5\,\mathrm{m}. The radius is measured to be 8\,\mathrm{m} with a possible error of \pm0.25\,\mathrm{m}. Use differentials to estimate the maximum error in the volume. Find the approximate relative error and the approximate percentage error.
Solution
Given
  • h = 5\,\mathrm{m} height (exact)
  • r = 8\,\mathrm{m} measured radius
  • dr = \pm0.25\,\mathrm{m} possible error in the radius
To find
  • maximum error dV
  • relative error
  • percentage error
  1. Volume of a cylinder with the fixed height h=5.
    V=\pi r^{2}h=5\pi r^{2}
  2. Differential of the volume.
    dV=10\pi r\,dr
  3. Substitute r=8 and dr=\pm0.25.
    dV=10\pi(8)(\pm0.25)=\pm20\pi\approx\pm62.83\,\mathrm{m^{3}}
  4. Volume for r=8.
    V=5\pi(8)^{2}=320\pi\approx1005.3\,\mathrm{m^{3}}
  5. Relative error.
    \dfrac{dV}{V}=\dfrac{\pm20\pi}{320\pi}=\pm\dfrac{1}{16}=\pm0.0625
  6. Percentage error.
    \pm0.0625\times100\%=\pm6.25\%
dV=\pm20\pi\approx\pm62.8\,\mathrm{m^{3}},\qquad \dfrac{dV}{V}=\pm0.0625,\qquad \pm6.25\%
8
A 15\,\mathrm{ft} ladder is leaning against a wall of a house. The bottom of the ladder is pulled away from the base of the wall at a constant rate of 2\,\mathrm{ft/min}. At what rate is the top of the ladder sliding down the wall when the bottom of the ladder is 5\,\mathrm{ft} from the wall?
Solution
The ladder, wall and ground form a right triangle with hypotenuse 15\,\mathrm{ft}.
  1. Let x be the distance of the foot from the wall and y the height of the top. The ladder is the hypotenuse.
    x^{2}+y^{2}=15^{2}=225
  2. Differentiate w.r.t. time t and solve for \frac{dy}{dt}.
    2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0\Rightarrow\dfrac{dy}{dt}=-\dfrac{x}{y}\dfrac{dx}{dt}
  3. When x=5:
    y=\sqrt{225-25}=\sqrt{200}=10\sqrt2
  4. Substitute \frac{dx}{dt}=2.
    \dfrac{dy}{dt}=-\dfrac{5}{10\sqrt2}(2)=-\dfrac{1}{\sqrt2}=-\dfrac{\sqrt2}{2}\approx-0.707\,\mathrm{ft/min}
  5. The minus sign means y is decreasing: the top slides down the wall at about 0.71\,\mathrm{ft/min}.
\dfrac{dy}{dt}=-\dfrac{1}{\sqrt2}\approx-0.707\,\mathrm{ft/min}\ (\text{sliding down at }0.71\,\mathrm{ft/min})
9
Find the absolute extrema of f(x)=x^{3}-3x^{2}-24x+2 on the given interval.
Solution
  1. Differentiate and factor.
    f'(x)=3x^{2}-6x-24=3(x^{2}-2x-8)=3(x-4)(x+2)
  2. Critical values: x=-2 and x=4. The function is a polynomial, so it is continuous and the closed interval method applies.
(a) [-3,1]
  1. Only x=-2 lies in [-3,1]. Evaluate f there and at the end points.
    \begin{aligned}f(-3)&=-27-27+72+2=20\\ f(-2)&=-8-12+48+2=30\\ f(1)&=1-3-24+2=-24\end{aligned}
  2. Compare: the largest value is 30 and the smallest is -24.
\text{absolute max}=f(-2)=30,\qquad \text{absolute min}=f(1)=-24
(b) [-3,8]
  1. Both x=-2 and x=4 lie in [-3,8]. Evaluate f at them and at the end points.
    \begin{aligned}f(-3)&=20\\ f(-2)&=30\\ f(4)&=64-48-96+2=-78\\ f(8)&=512-192-192+2=130\end{aligned}
  2. Compare: the largest value is 130 and the smallest is -78.
\text{absolute max}=f(8)=130,\qquad \text{absolute min}=f(4)=-78
10
Graph the function f(x)=x+\dfrac{1}{x}.
Solution
  1. Domain. x\neq0, because of \frac{1}{x}.
  2. Symmetry. f(-x)=-x-\frac{1}{x}=-f(x), so f is odd and the graph is symmetric about the origin.
  3. Intercepts. x=0 is excluded, so there is no y-intercept. x+\frac1x=0 gives x^{2}=-1, which is impossible, so there is no x-intercept.
  4. Asymptotes. As x\to0^{+}, f(x)\to+\infty and as x\to0^{-}, f(x)\to-\infty, so x=0 is a vertical asymptote. Also f(x)-x=\frac1x\to0 as x\to\pm\infty, so the graph gets closer and closer to the line y=x.
  5. Critical values. f'(x)=0 gives x^{2}=1, so x=\pm1.
    f'(x)=1-\dfrac{1}{x^{2}}=\dfrac{x^{2}-1}{x^{2}}
  6. Increasing and decreasing. f'(x)>0 when x^{2}>1 and f'(x)<0 when 0<x^{2}<1.
    \text{increasing on }(-\infty,-1)\text{ and }(1,\infty);\qquad \text{decreasing on }(-1,0)\text{ and }(0,1)
  7. Second derivative test. f''(1)=2>0 gives a relative minimum f(1)=2; f''(-1)=-2<0 gives a relative maximum f(-1)=-2.
    f''(x)=\dfrac{2}{x^{3}}
  8. Concavity. f''(x)<0 for x<0 (concave down) and f''(x)>0 for x>0 (concave up). There is no point of inflection because x=0 is not in the domain.
  9. Table of values (the values for negative x are the negatives of these).
    \begin{array}{c|cccccc} x & \tfrac14 & \tfrac12 & 1 & 2 & 3 & 4\\ \hline f(x) & 4.25 & 2.5 & 2 & 2.5 & 3.33 & 4.25\end{array}
  10. Plot the points, draw the asymptotes x=0 and y=x, and join the points with two smooth branches.
Two branches: the right branch has its lowest point at (1,2), the left branch its highest point at (-1,-2). The range is (-\infty,-2]\cup[2,\infty).
\text{rel. max }(-1,-2),\ \text{rel. min }(1,2);\ \text{asymptotes }x=0,\ y=x;\ \text{range }(-\infty,-2]\cup[2,\infty)

FBISE 2026 questions

Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 21 mark
If f(x)=\ln(2x), then f'(3) equals:
  1. A\dfrac13Answer
  2. B\dfrac23
  3. C\dfrac16
  4. D\dfrac32
Why

By the chain rule f'(x)=\dfrac{1}{2x}\cdot2=\dfrac1x (or write \ln2x=\ln2+\ln x). So f'(3)=\dfrac13. Option B forgets that the factor 2 cancels.

2026 Annual I, Set B, MCQ 91 mark
For f(x)=x^{3}-3x^{2}+2, the value of f''(2) gives:
  1. Alocal maximum
  2. Blocal minimumAnswer
  3. Cinflection point
  4. Dcannot be determined
Why

f'(x)=3x^{2}-6x, so f'(2)=0 and x=2 is a critical value. f''(x)=6x-6 gives f''(2)=6>0, so by the second derivative test f(2)=-2 is a local minimum. The inflection point is at x=1, where f''=0.

2026 Annual I, Set B, MCQ 191 mark
A side of a cube is 4\,\mathrm{cm}. If the side increases by 0.1\,\mathrm{cm}, the approximate increase in volume is:
  1. A1.2
  2. B2.4
  3. C3.6
  4. D4.8Answer
Why

V=x^{3}, so dV=3x^{2}\,dx=3(4)^{2}(0.1)=4.8\,\mathrm{cm^{3}}. (The exact increase is 4.1^{3}-4^{3}=4.921\,\mathrm{cm^{3}}, close to the differential estimate.)

2026 Annual I, Set B, Q.2 (ii)4 marks
Find \dfrac{dy}{dx} for y=e^{2x}\ln(x).
Solution
  1. y is a product of e^{2x} and \ln x, so use the product rule.
    \dfrac{dy}{dx}=e^{2x}\dfrac{d}{dx}(\ln x)+\ln x\,\dfrac{d}{dx}\left(e^{2x}\right)
  2. Differentiate each factor: \dfrac{d}{dx}\ln x=\dfrac1x and, by the chain rule, \dfrac{d}{dx}e^{2x}=2e^{2x}.
    \dfrac{dy}{dx}=e^{2x}\cdot\dfrac1x+2e^{2x}\ln x
  3. Take out the common factor e^{2x} and write over one denominator (x>0).
    \dfrac{dy}{dx}=e^{2x}\left(\dfrac1x+2\ln x\right)=\dfrac{e^{2x}(1+2x\ln x)}{x}
\dfrac{dy}{dx}=\dfrac{e^{2x}(1+2x\ln x)}{x}
2026 Annual I, Set B, Q.2 (iii) OR4 marks
Determine the intervals in which f(x)=x^{3}-3x^{2}-9x+5 is increasing or decreasing.
Solution
  1. Differentiate and factorise.
    f'(x)=3x^{2}-6x-9=3(x^{2}-2x-3)=3(x+1)(x-3)
  2. Solve f'(x)=0 to find where the sign can change.
    x=-1\quad\text{or}\quad x=3
  3. Signs of f': for x<-1 both factors are negative, so f'>0; for -1<x<3 the factors have opposite signs, so f'<0; for x>3 both are positive, so f'>0.
IntervalTest valuef'(x)Behaviour
(-\infty,-1)f'(-2)=15+increasing
(-1,3)f'(0)=-9-decreasing
(3,\infty)f'(4)=15+increasing
Turning points: local maximum 10 at x = -1 and local minimum -22 at x = 3.
\text{increasing on }(-\infty,-1)\cup(3,\infty),\quad\text{decreasing on }(-1,3)
2026 Annual I, Set B, Q.2 (viii)4 marks
The radius of a sphere is measured as 10\,\mathrm{cm} with a possible error of 0.05\,\mathrm{cm}. Use differentials to find the approximate percentage error in the volume.
Solution
Given
  • r = 10\,\mathrm{cm} measured radius
  • dr = 0.05\,\mathrm{cm} possible error
To find
  • percentage error in V
  1. Volume of a sphere and its differential.
    V=\dfrac43\pi r^{3}\ \Longrightarrow\ dV=4\pi r^{2}\,dr
  2. Approximate error in the volume.
    dV=4\pi(10)^{2}(0.05)=20\pi\approx62.8\,\mathrm{cm^{3}}
  3. Volume itself.
    V=\dfrac43\pi(10)^{3}=\dfrac{4000\pi}{3}\approx4188.8\,\mathrm{cm^{3}}
  4. Relative error, then percentage. (Shortcut: \dfrac{dV}{V}=3\dfrac{dr}{r}.)
    \dfrac{dV}{V}\times100\%=\dfrac{20\pi}{4000\pi/3}\times100\%=3\left(\dfrac{0.05}{10}\right)\times100\%=1.5\%
\text{percentage error}\approx1.5\%
2026 Annual I, Set B, Q.2 (xi) OR4 marks
Find the derivative of f(x)=\dfrac{x^{2}+1}{x^{2}-1} and simplify.
Solution
  1. Quotient rule: denominator times the derivative of the numerator, minus numerator times the derivative of the denominator, over the denominator squared.
    f'(x)=\dfrac{(x^{2}-1)(2x)-(x^{2}+1)(2x)}{(x^{2}-1)^{2}}
  2. Take out the common factor 2x in the numerator.
    =\dfrac{2x\left[(x^{2}-1)-(x^{2}+1)\right]}{(x^{2}-1)^{2}}
  3. Simplify the bracket: x^{2}-1-x^{2}-1=-2.
    =\dfrac{-4x}{(x^{2}-1)^{2}},\qquad x\neq\pm1
f'(x)=-\dfrac{4x}{(x^{2}-1)^{2}}
2026 Annual I, Set B, Q.38 marks
Let f(x)=\begin{cases}ax^{2}+bx+2, & x\le1\\ 3x+a, & 1<x<3\\ \dfrac{2x^{2}+b}{x}, & x\ge3\end{cases}. Given that f(x) is continuous at x=1 and at x=3, find the values of a and b.
Solution

Condition for continuity

f is continuous at x=c when f(c) is defined, \lim_{x\to c}f(x) exists, and \lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=f(c). Each piece is a polynomial or a rational function whose denominator is non-zero on its interval, so only the joining points x=1 and x=3 need to be checked.

Continuity at x = 1

  1. The first piece contains x=1 (condition x\le1), so it gives both f(1) and the left-hand limit.
    f(1)=\lim_{x\to1^-}(ax^{2}+bx+2)=a+b+2
  2. The middle piece holds just to the right of 1.
    \lim_{x\to1^+}(3x+a)=3+a
  3. Equate them.
    a+b+2=3+a\ \Longrightarrow\ b=1\qquad(1)

Continuity at x = 3

  1. The middle piece holds just to the left of 3.
    \lim_{x\to3^-}(3x+a)=9+a
  2. The last piece contains x=3 (condition x\ge3), so it gives f(3) and the right-hand limit.
    f(3)=\lim_{x\to3^+}\dfrac{2x^{2}+b}{x}=\dfrac{18+b}{3}
  3. Equate them and multiply by 3.
    9+a=\dfrac{18+b}{3}\ \Longrightarrow\ 27+3a=18+b\qquad(2)

Solving

  1. Substitute b=1 from (1) into (2).
    27+3a=19\ \Longrightarrow\ 3a=-8\ \Longrightarrow\ a=-\dfrac83

Check

  1. At x=1: a+b+2=-\tfrac83+1+2=\tfrac13 and 3+a=3-\tfrac83=\tfrac13. At x=3: 9+a=\tfrac{19}{3} and \dfrac{18+1}{3}=\tfrac{19}{3}. Both joins match, so f is continuous at x=1 and x=3.
With a=-\tfrac83 and b=1 the three pieces join without a break.
a=-\dfrac83,\qquad b=1
2026 Annual I, Set B, Q.48 marks
Given the function f(x)=2\sin x+\cos2x, x\in[0,\pi]. (a) Find all critical points of f(x). (b) Use the second derivative test to classify each critical point as a local maximum or local minimum. (c) Determine the absolute (global) maximum and absolute (global) minimum values of f(x).
Solution

(a) Critical points

Critical points are the numbers in [0,\pi] where f'(x)=0 or f' does not exist. Here f' exists everywhere, so we solve f'(x)=0.

  1. Differentiate; use \sin2x=2\sin x\cos x to factorise.
    f'(x)=2\cos x-2\sin2x=2\cos x-4\sin x\cos x=2\cos x\,(1-2\sin x)
  2. Set each factor to zero and keep the solutions in [0,\pi].
    \cos x=0\Rightarrow x=\dfrac{\pi}{2};\qquad\sin x=\dfrac12\Rightarrow x=\dfrac{\pi}{6},\ \dfrac{5\pi}{6}
  3. Values of f at these points.
    f\!\left(\tfrac{\pi}{6}\right)=1+\tfrac12=\tfrac32,\quad f\!\left(\tfrac{\pi}{2}\right)=2-1=1,\quad f\!\left(\tfrac{5\pi}{6}\right)=1+\tfrac12=\tfrac32

(b) Second derivative test

  1. Differentiate again.
    f''(x)=-2\sin x-4\cos2x
  2. At x=\tfrac{\pi}{6}: \sin x=\tfrac12, \cos2x=\cos\tfrac{\pi}{3}=\tfrac12.
    f''\!\left(\tfrac{\pi}{6}\right)=-1-2=-3<0\ \Longrightarrow\ \text{local maximum}
  3. At x=\tfrac{\pi}{2}: \sin x=1, \cos2x=\cos\pi=-1.
    f''\!\left(\tfrac{\pi}{2}\right)=-2+4=2>0\ \Longrightarrow\ \text{local minimum}
  4. At x=\tfrac{5\pi}{6}: \sin x=\tfrac12, \cos2x=\cos\tfrac{5\pi}{3}=\tfrac12.
    f''\!\left(\tfrac{5\pi}{6}\right)=-1-2=-3<0\ \Longrightarrow\ \text{local maximum}

(c) Absolute extrema

f is continuous on the closed interval [0,\pi], so it has an absolute maximum and minimum (extreme value theorem). Compare the values at the critical points and at the endpoints.

x0\tfrac{\pi}{6}\tfrac{\pi}{2}\tfrac{5\pi}{6}\pi
f(x)1\tfrac321\tfrac321

The largest value in the table is \tfrac32 and the smallest is 1. So the absolute maximum value of f is \tfrac32, reached at both x=\tfrac{\pi}{6} and x=\tfrac{5\pi}{6}, and the absolute minimum value is 1, reached at x=0, x=\tfrac{\pi}{2} and x=\pi. Here the local minimum at \tfrac{\pi}{2} happens to equal the endpoint values.

\text{abs. max}=\tfrac32\ \text{at }x=\tfrac{\pi}{6},\tfrac{5\pi}{6};\qquad\text{abs. min}=1\ \text{at }x=0,\tfrac{\pi}{2},\pi
2026 Annual I, Set D, MCQ 31 mark
If f(x)=\ln(4x+1), then f'(2) equals:
  1. A9
  2. B\dfrac49Answer
  3. C\dfrac12
  4. D\dfrac19
Why

f'(x)=\dfrac{4}{4x+1} (chain rule: derivative of the inside over the inside). At x=2: \dfrac{4}{9}. Option D forgets the factor 4 from the inside.

2026 Annual I, Set D, MCQ 101 mark
The critical points of f(x)=x^{3}-3x^{2}+2 on the closed interval [-1,3] are:
  1. Ax=-1,0,2
  2. Bx=0,2 onlyAnswer
  3. Cx=-1,0,2,3
  4. Dx=0,2,3
Why

f'(x)=3x^{2}-6x=3x(x-2) exists everywhere and is zero at x=0 and x=2, both inside [-1,3]. The endpoints -1 and 3 are not critical points; they are checked separately when finding absolute extrema, which is why options A, C and D are tempting but wrong.

2026 Annual I, Set D, MCQ 201 mark
The radius of a sphere is 5\,\mathrm{cm}. If the radius increases by 0.2\,\mathrm{cm}, the approximate increase in volume is:
  1. A25\pi\,\mathrm{cm^{3}}
  2. B10\pi\,\mathrm{cm^{3}}
  3. C15\pi\,\mathrm{cm^{3}}
  4. D20\pi\,\mathrm{cm^{3}}Answer
Why

V=\dfrac43\pi r^{3} gives dV=4\pi r^{2}\,dr=4\pi(25)(0.2)=20\pi\,\mathrm{cm^{3}}.

2026 Annual I, Set D, Q.2 (ii)4 marks
Find \dfrac{dy}{dx} if y=e^{3x}\ln\left(\dfrac{2x}{x^{2}+1}\right).
Solution
  1. Product rule with first factor e^{3x} and second factor \ln\dfrac{2x}{x^{2}+1}.
    \dfrac{dy}{dx}=e^{3x}\dfrac{d}{dx}\left[\ln\dfrac{2x}{x^{2}+1}\right]+\ln\left(\dfrac{2x}{x^{2}+1}\right)\dfrac{d}{dx}e^{3x}
  2. Simplify the logarithm with the log laws before differentiating (x>0).
    \ln\dfrac{2x}{x^{2}+1}=\ln2+\ln x-\ln(x^{2}+1)
  3. Differentiate it term by term and combine.
    \dfrac{d}{dx}\ln\dfrac{2x}{x^{2}+1}=0+\dfrac1x-\dfrac{2x}{x^{2}+1}=\dfrac{x^{2}+1-2x^{2}}{x(x^{2}+1)}=\dfrac{1-x^{2}}{x(x^{2}+1)}
  4. Also \dfrac{d}{dx}e^{3x}=3e^{3x}. Substitute and take out e^{3x}.
    \dfrac{dy}{dx}=e^{3x}\left[\dfrac{1-x^{2}}{x(x^{2}+1)}+3\ln\left(\dfrac{2x}{x^{2}+1}\right)\right]
\dfrac{dy}{dx}=e^{3x}\left[3\ln\dfrac{2x}{x^{2}+1}+\dfrac{1-x^{2}}{x(x^{2}+1)}\right]
2026 Annual I, Set D, Q.2 (iii) OR4 marks
Determine the intervals in which the function f(x)=x^{3}-6x^{2}+9x-4 is increasing or decreasing.
Solution
  1. Differentiate and factorise.
    f'(x)=3x^{2}-12x+9=3(x^{2}-4x+3)=3(x-1)(x-3)
  2. Solve f'(x)=0.
    x=1\quad\text{or}\quad x=3
  3. Test the sign of f' in each interval.
IntervalTest valuef'(x)Behaviour
(-\infty,1)f'(0)=9+increasing
(1,3)f'(2)=-3-decreasing
(3,\infty)f'(4)=9+increasing
Local maximum 0 at x = 1 and local minimum -4 at x = 3.
\text{increasing on }(-\infty,1)\cup(3,\infty),\quad\text{decreasing on }(1,3)
2026 Annual I, Set D, Q.2 (viii)4 marks
The side length of a cube is measured as 12\,\mathrm{cm} with a possible error of 0.03\,\mathrm{cm}. Use differentials to find the approximate percentage error in the volume.
Solution
Given
  • x = 12\,\mathrm{cm} measured side
  • dx = 0.03\,\mathrm{cm} possible error
To find
  • percentage error in V
  1. Volume and its differential.
    V=x^{3}\ \Longrightarrow\ dV=3x^{2}\,dx
  2. Approximate error in the volume.
    dV=3(12)^{2}(0.03)=12.96\,\mathrm{cm^{3}}
  3. The volume is 12^{3}=1728\,\mathrm{cm^{3}}. Percentage error (shortcut \dfrac{dV}{V}=3\dfrac{dx}{x}).
    \dfrac{dV}{V}\times100\%=\dfrac{12.96}{1728}\times100\%=3\left(\dfrac{0.03}{12}\right)\times100\%=0.75\%
\text{percentage error}\approx0.75\%
2026 Annual I, Set D, Q.2 (xi) OR4 marks
Differentiate the function w.r.t. x: y=\dfrac{2x^{3}-3x+1}{x^{2}+1}.
Solution
  1. Quotient rule with numerator 2x^{3}-3x+1 (derivative 6x^{2}-3) and denominator x^{2}+1 (derivative 2x).
    \dfrac{dy}{dx}=\dfrac{(x^{2}+1)(6x^{2}-3)-(2x^{3}-3x+1)(2x)}{(x^{2}+1)^{2}}
  2. Expand both products.
    \begin{aligned}(x^{2}+1)(6x^{2}-3)&=6x^{4}+3x^{2}-3\\ (2x^{3}-3x+1)(2x)&=4x^{4}-6x^{2}+2x\end{aligned}
  3. Subtract and collect like terms.
    \dfrac{dy}{dx}=\dfrac{2x^{4}+9x^{2}-2x-3}{(x^{2}+1)^{2}}
\dfrac{dy}{dx}=\dfrac{2x^{4}+9x^{2}-2x-3}{(x^{2}+1)^{2}}
2026 Annual I, Set D, Q.38 marks
Let f(x)=\begin{cases}\dfrac m2x^{2}-\dfrac n3x+1, & x\le2\\ \dfrac{3x+m}{4}, & 2<x<5\\ \dfrac{x^{2}+2n}{x+1}, & x\ge5\end{cases}. Given that f(x) is continuous at x=2 and at x=5, find the values of m and n.
Solution

Condition for continuity

f is continuous at x=c when \lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=f(c). The pieces are continuous on their own intervals (the last denominator x+1 is not zero for x\ge5), so we only need the joins at x=2 and x=5.

Continuity at x = 2

  1. The first piece contains x=2, so it gives f(2) and the left-hand limit.
    f(2)=\dfrac m2(4)-\dfrac n3(2)+1=2m-\dfrac{2n}{3}+1
  2. The middle piece gives the right-hand limit.
    \lim_{x\to2^+}\dfrac{3x+m}{4}=\dfrac{6+m}{4}
  3. Equate and multiply by 12 to clear fractions.
    24m-8n+12=18+3m\ \Longrightarrow\ 21m-8n=6\qquad(1)

Continuity at x = 5

  1. The middle piece gives the left-hand limit.
    \lim_{x\to5^-}\dfrac{3x+m}{4}=\dfrac{15+m}{4}
  2. The last piece contains x=5, so it gives f(5) and the right-hand limit.
    f(5)=\dfrac{25+2n}{6}
  3. Equate and cross-multiply.
    6(15+m)=4(25+2n)\ \Longrightarrow\ 90+6m=100+8n\ \Longrightarrow\ 6m-8n=10\qquad(2)

Solving

  1. Subtract (2) from (1) to eliminate n.
    15m=-4\ \Longrightarrow\ m=-\dfrac{4}{15}
  2. Substitute in (2).
    8n=6m-10=-\dfrac{8}{5}-10=-\dfrac{58}{5}\ \Longrightarrow\ n=-\dfrac{29}{20}

Check

  1. At x=2: 2m-\tfrac{2n}{3}+1=-\tfrac{8}{15}+\tfrac{29}{30}+1=\tfrac{43}{30} and \tfrac{6+m}{4}=\tfrac{86/15}{4}=\tfrac{43}{30}. At x=5: \tfrac{15+m}{4}=\tfrac{221}{60} and \tfrac{25+2n}{6}=\tfrac{22.1}{6}=\tfrac{221}{60}. Both joins match.
m=-\dfrac{4}{15},\qquad n=-\dfrac{29}{20}
2026 Annual I, Set D, Q.48 marks
Given the function: f(x)=3\sin x+\cos2x, x\in[0,\pi]. (a) Find all critical points of f(x). (b) Use the second derivative test to classify each critical point as a local maximum or local minimum. (c) Determine the absolute (global) maximum and absolute (global) minimum values of f(x).
Solution

(a) Critical points

f' exists for every x, so the critical points are the solutions of f'(x)=0 in [0,\pi].

  1. Differentiate and factorise using \sin2x=2\sin x\cos x.
    f'(x)=3\cos x-2\sin2x=3\cos x-4\sin x\cos x=\cos x\,(3-4\sin x)
  2. \cos x=0 gives one solution in [0,\pi].
    x=\dfrac{\pi}{2}
  3. \sin x=\tfrac34 gives two solutions in [0,\pi] (first and second quadrants).
    x=\alpha=\sin^{-1}\tfrac34\approx0.848\ (48.6^{\circ}),\qquad x=\pi-\alpha\approx2.294\ (131.4^{\circ})
  4. Values of f. When \sin x=\tfrac34, \cos2x=1-2\sin^{2}x=1-\tfrac98=-\tfrac18.
    f(\alpha)=f(\pi-\alpha)=\tfrac94-\tfrac18=\tfrac{17}{8},\qquad f\!\left(\tfrac{\pi}{2}\right)=3+\cos\pi=2

(b) Second derivative test

  1. Differentiate again.
    f''(x)=-3\sin x-4\cos2x
  2. At x=\alpha and x=\pi-\alpha: \sin x=\tfrac34 and \cos2x=-\tfrac18.
    f''=-\tfrac94+\tfrac12=-\tfrac74<0\ \Longrightarrow\ \text{local maximum at both}
  3. At x=\tfrac{\pi}{2}: \sin x=1, \cos2x=-1.
    f''\!\left(\tfrac{\pi}{2}\right)=-3+4=1>0\ \Longrightarrow\ \text{local minimum}

(c) Absolute extrema

f is continuous on [0,\pi], so compare the critical values with the endpoint values f(0)=0+1=1 and f(\pi)=0+\cos2\pi=1.

x0\alpha\approx0.848\tfrac{\pi}{2}\pi-\alpha\approx2.294\pi
f(x)1\tfrac{17}{8}2\tfrac{17}{8}1

So the absolute maximum value is \tfrac{17}{8}, reached twice, at x=\sin^{-1}\tfrac34 and at x=\pi-\sin^{-1}\tfrac34. The absolute minimum value is 1, at the endpoints x=0 and x=\pi. The local minimum 2 at x=\tfrac{\pi}{2} is not the absolute minimum, because the endpoints give the smaller value 1.

\text{abs. max}=\tfrac{17}{8}\ \text{at }x=\sin^{-1}\tfrac34,\ \pi-\sin^{-1}\tfrac34;\qquad\text{abs. min}=1\ \text{at }x=0,\ \pi
2026 Annual I, Old syllabus (Curriculum 2000), MCQ 2, old syllabus1 mark
If f(x)=x^{4}, then the second derivative f''(x) is:
  1. A4x^{3}
  2. B12x^{2}Answer
  3. C24x
  4. D24
Why

f'(x)=4x^{3} and f''(x)=12x^{2}. Option A is only the first derivative; C and D are the third and fourth derivatives.

2026 Annual I, Old syllabus (Curriculum 2000), MCQ 9, old syllabus1 mark
If y=\sin(3x^{2}), then \dfrac{dy}{dx} is:
  1. A\cos(3x^{2})
  2. B6x\cos(3x^{2})Answer
  3. C\cos(6x)
  4. D3x^{2}\cos(3x^{2})
Why

Chain rule: the derivative of the outer \sin is \cos(3x^{2}), times the derivative of the inside, 6x. Option A forgets the inside derivative.

2026 Annual I, Old syllabus (Curriculum 2000), MCQ 16, old syllabus1 mark
The function f(x)=x^{2}-4x+3 has a minimum value at:
  1. Ax=0
  2. Bx=1
  3. Cx=2Answer
  4. Dx=3
Why

f'(x)=2x-4=0 gives x=2, and f''(x)=2>0, so the minimum is at x=2 (the minimum value is f(2)=-1). x=1 and x=3 are the zeros of f, not its minimum.

2026 Annual I, Old syllabus (Curriculum 2000), Q.2 (ii), old syllabus4 marks
Determine whether f(x)=-4x^{3}+6x^{2}+2 is increasing or decreasing at x=1.
Solution
  1. The sign of f' at a point decides: f'>0 increasing, f'<0 decreasing.
    f'(x)=-12x^{2}+12x=-12x(x-1)
  2. Evaluate at x=1.
    f'(1)=-12+12=0
  3. Since f'(1)=0, f is neither increasing nor decreasing at x=1: the tangent there is horizontal (a stationary point).
  4. To describe the point fully, use the second derivative.
    f''(x)=-24x+12,\qquad f''(1)=-12<0\ \Longrightarrow\ \text{local maximum},\ f(1)=4
  5. So f is increasing just before x=1 (on (0,1), where f'>0) and decreasing just after it (on (1,\infty), where f'<0).
f'(1)=0:\ \text{neither increasing nor decreasing at }x=1\ (\text{local maximum})

Note. The question expects the sign of f'(1) to be tested; because f'(1)=0, the correct conclusion is that x=1 is a stationary point (a local maximum), not a point where f is increasing or decreasing.

2026 Annual I, Old syllabus (Curriculum 2000), Q.2 (vi) OR, old syllabus4 marks
Evaluate:
Solution
(a) \lim_{x\to\infty}\left(1+\dfrac2x\right)^{3x}
  1. Use \lim_{u\to\infty}\left(1+\dfrac1u\right)^{u}=e. Put \dfrac2x=\dfrac1u, so x=2u and u\to\infty as x\to\infty.
    \left(1+\dfrac2x\right)^{3x}=\left(1+\dfrac1u\right)^{6u}=\left[\left(1+\dfrac1u\right)^{u}\right]^{6}
  2. Take the limit.
    \lim_{x\to\infty}\left(1+\dfrac2x\right)^{3x}=e^{6}
e^{6}
(b) \lim_{x\to0}(1+4x)^{1/x}
  1. Use \lim_{u\to0}(1+u)^{1/u}=e. Put u=4x, so \dfrac1x=\dfrac4u and u\to0.
    (1+4x)^{1/x}=\left[(1+u)^{1/u}\right]^{4}
  2. Take the limit.
    \lim_{x\to0}(1+4x)^{1/x}=e^{4}
e^{4}

Note. These e-type limits are not in the new book's text; the notes on important limits cover them.

2026 Annual I, Old syllabus (Curriculum 2000), Q.2 (vii) OR, old syllabus4 marks
Given f(x)=\ln(2x+1), find f'(x), f''(x), f'''(x) and f^{(4)}(x).
Solution
  1. Chain rule: derivative of the inside over the inside.
    f'(x)=\dfrac{2}{2x+1}=2(2x+1)^{-1}
  2. Power rule for functions; each step brings down the power and a factor 2 from the inside.
    f''(x)=2(-1)(2x+1)^{-2}(2)=-\dfrac{4}{(2x+1)^{2}}
  3. f'''(x)=-4(-2)(2x+1)^{-3}(2)=\dfrac{16}{(2x+1)^{3}}
  4. f^{(4)}(x)=16(-3)(2x+1)^{-4}(2)=-\dfrac{96}{(2x+1)^{4}}
f'=\dfrac{2}{2x+1},\ f''=-\dfrac{4}{(2x+1)^{2}},\ f'''=\dfrac{16}{(2x+1)^{3}},\ f^{(4)}=-\dfrac{96}{(2x+1)^{4}}
2026 Annual I, Old syllabus (Curriculum 2000), Q.2 (x), old syllabus4 marks
Use differentials to find the approximate value of \sin46^{\circ}.
Solution
  1. Take f(x)=\sin x with the known angle x=45^{\circ}=\dfrac{\pi}{4} and the change dx=1^{\circ}, converted to radians.
    dx=1^{\circ}=\dfrac{\pi}{180}\approx0.017453
  2. Linear approximation f(x+dx)\approx f(x)+f'(x)\,dx with f'(x)=\cos x.
    \sin46^{\circ}\approx\sin45^{\circ}+\cos45^{\circ}\cdot\dfrac{\pi}{180}
  3. Substitute \sin45^{\circ}=\cos45^{\circ}=\dfrac{1}{\sqrt2}\approx0.70711.
    \sin46^{\circ}\approx0.70711+0.70711(0.017453)=0.70711+0.01234=0.71945
\sin46^{\circ}\approx0.7194
2026 Annual I, Old syllabus (Curriculum 2000), Q.2 (xi) OR, old syllabus4 marks
A curve is defined parametrically by x=e^{t}\cos t, y=e^{t}\sin t. Find \dfrac{dy}{dx} in terms of t.
Solution
  1. Differentiate x with the product rule.
    \dfrac{dx}{dt}=e^{t}\cos t-e^{t}\sin t=e^{t}(\cos t-\sin t)
  2. Differentiate y with the product rule.
    \dfrac{dy}{dt}=e^{t}\sin t+e^{t}\cos t=e^{t}(\sin t+\cos t)
  3. Divide; e^{t} cancels.
    \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{\sin t+\cos t}{\cos t-\sin t}
  4. Optional: divide the top and bottom by \cos t.
    \dfrac{dy}{dx}=\dfrac{1+\tan t}{1-\tan t}=\tan\left(\dfrac{\pi}{4}+t\right)
\dfrac{dy}{dx}=\dfrac{\cos t+\sin t}{\cos t-\sin t}
2026 Annual I, Old syllabus (Curriculum 2000), Q.3, old syllabus8 marks
Test the continuity of the function f(x)=\begin{cases}\dfrac{x^{2}-4}{x-2}, & x<2\\ 4, & 2\le x<5\\ x^{2}-6x+9, & x\ge5\end{cases} at x=2 and x=5.
Solution

Condition for continuity

f is continuous at x=c if (i) f(c) is defined, (ii) \lim_{x\to c}f(x) exists, that is, the left-hand and right-hand limits are equal, and (iii) this limit equals f(c). We test each condition at x=2 and at x=5, using the piece that applies on each side.

At x = 2

  1. (i) x=2 satisfies 2\le x<5, so the middle piece gives the value.
    f(2)=4
  2. Left-hand limit: for x<2 the first piece applies; cancel x-2 (allowed since x\neq2).
    \lim_{x\to2^-}\dfrac{(x-2)(x+2)}{x-2}=\lim_{x\to2^-}(x+2)=4
  3. Right-hand limit: the middle piece is the constant 4.
    \lim_{x\to2^+}4=4
  4. (ii) The one-sided limits agree, so the limit exists; (iii) it equals f(2).
    \lim_{x\to2}f(x)=4=f(2)\ \Longrightarrow\ f\ \text{is continuous at }x=2

At x = 5

  1. (i) x=5 satisfies x\ge5, so the last piece gives the value.
    f(5)=25-30+9=4
  2. Left-hand limit from the middle piece.
    \lim_{x\to5^-}4=4
  3. Right-hand limit from the last piece.
    \lim_{x\to5^+}(x^{2}-6x+9)=\lim_{x\to5^+}(x-3)^{2}=4
  4. All three agree.
    \lim_{x\to5}f(x)=4=f(5)\ \Longrightarrow\ f\ \text{is continuous at }x=5

Conclusion

f is continuous at both x=2 and x=5. The graph below shows the three pieces joining without a gap: the line y=x+2 (with x=2 excluded) reaches height 4, the horizontal segment stays at 4, and the parabola (x-3)^{2} starts from height 4 at x=5.

f\ \text{is continuous at }x=2\ \text{and at }x=5
2026 Annual I, Old syllabus (Curriculum 2000), Q.4, old syllabus8 marks
Differentiate the following w.r.t. x: y=4\sin^{-1}\left(\dfrac x4\right)+\cos^{-1}\left(\dfrac x4\right)+2\tan^{-1}\left(\dfrac x2\right)+5\cot^{-1}\left(\dfrac x2\right)
Solution

Formulas used

With u a function of x: \dfrac{d}{dx}\sin^{-1}u=\dfrac{u'}{\sqrt{1-u^{2}}}, \dfrac{d}{dx}\cos^{-1}u=-\dfrac{u'}{\sqrt{1-u^{2}}}, \dfrac{d}{dx}\tan^{-1}u=\dfrac{u'}{1+u^{2}} and \dfrac{d}{dx}\cot^{-1}u=-\dfrac{u'}{1+u^{2}}. Here u=\dfrac x4 (so u'=\dfrac14) in the first two terms and u=\dfrac x2 (so u'=\dfrac12) in the last two. The answer is valid for -4<x<4.

Term by term

  1. First term. Simplify \sqrt{1-\dfrac{x^{2}}{16}}=\dfrac{\sqrt{16-x^{2}}}{4}.
    \dfrac{d}{dx}\left[4\sin^{-1}\dfrac x4\right]=4\cdot\dfrac{1/4}{\sqrt{1-x^{2}/16}}=\dfrac{4}{\sqrt{16-x^{2}}}
  2. Second term (the same root, with a minus sign).
    \dfrac{d}{dx}\left[\cos^{-1}\dfrac x4\right]=-\dfrac{1/4}{\sqrt{1-x^{2}/16}}=-\dfrac{1}{\sqrt{16-x^{2}}}
  3. Third term. Simplify 1+\dfrac{x^{2}}{4}=\dfrac{4+x^{2}}{4}.
    \dfrac{d}{dx}\left[2\tan^{-1}\dfrac x2\right]=2\cdot\dfrac{1/2}{1+x^{2}/4}=\dfrac{4}{4+x^{2}}
  4. Fourth term.
    \dfrac{d}{dx}\left[5\cot^{-1}\dfrac x2\right]=-5\cdot\dfrac{1/2}{1+x^{2}/4}=-\dfrac{10}{4+x^{2}}

Adding

\dfrac{dy}{dx}=\dfrac{4-1}{\sqrt{16-x^{2}}}+\dfrac{4-10}{4+x^{2}}=\dfrac{3}{\sqrt{16-x^{2}}}-\dfrac{6}{4+x^{2}}

The two \sin^{-1} and \cos^{-1} terms share the root \sqrt{16-x^{2}}, so they combine to \dfrac{3}{\sqrt{16-x^{2}}}; the \tan^{-1} and \cot^{-1} terms share the denominator 4+x^{2} and combine to -\dfrac{6}{4+x^{2}}.

Check with identities

Because \sin^{-1}u+\cos^{-1}u=\dfrac{\pi}{2} and \tan^{-1}v+\cot^{-1}v=\dfrac{\pi}{2}, the function can be rewritten as y=3\sin^{-1}\dfrac x4+3\cot^{-1}\dfrac x2+\dfrac{3\pi}{2}. Differentiating this shorter form gives \dfrac{3}{\sqrt{16-x^{2}}}-\dfrac{6}{4+x^{2}} again, which confirms the answer.

\dfrac{dy}{dx}=\dfrac{3}{\sqrt{16-x^{2}}}-\dfrac{6}{4+x^{2}}
2026 Annual I, Old syllabus (Curriculum 2006), MCQ 2, old syllabus1 mark
If \lim_{x\to3}g(x)=-2 and \lim_{x\to3}f(x)=4, then \lim_{x\to3}\dfrac{2f(x)-[g(x)]^{2}}{f(x)g(x)} results:
  1. A-\dfrac32
  2. B-\dfrac12Answer
  3. C\dfrac12
  4. D\dfrac52
Why

By the limit theorems (the denominator's limit 4(-2)=-8 is not zero): \dfrac{2(4)-(-2)^{2}}{4(-2)}=\dfrac{8-4}{-8}=-\dfrac12.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 3, old syllabus1 mark
What is the value of \left(\dfrac{dy}{dx}\right)_{x=1}, if y=x^{x}?
  1. A0
  2. B1Answer
  3. C1+\ln2
  4. D2
Why

Logarithmic differentiation: \ln y=x\ln x, so \dfrac1y\dfrac{dy}{dx}=1+\ln x and \dfrac{dy}{dx}=x^{x}(1+\ln x). At x=1: 1\cdot(1+0)=1.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 4, old syllabus1 mark
If y=\sinh x, then value of \left(\dfrac{dy}{dx}\right)_{x=0} is:
  1. A-1
  2. B0
  3. C1Answer
  4. Dundefined
Why

\dfrac{d}{dx}\sinh x=\cosh x=\dfrac{e^{x}+e^{-x}}{2}, so at x=0 the value is \dfrac{1+1}{2}=1. (Option B is \sinh0, the value of the function, not of its derivative.)

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 5, old syllabus1 mark
At what angle graphs of y=x and y=x^{2} intersect each other at origin?
  1. A0^{\circ}
  2. B30^{\circ}
  3. C45^{\circ}Answer
  4. D90^{\circ}
Why

The angle between curves is the angle between their tangents. At the origin the slope of y=x is 1 and the slope of y=x^{2} is 2x=0. So \tan\theta=\left|\dfrac{1-0}{1+1\cdot0}\right|=1 and \theta=45^{\circ}.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (ii), old syllabus4 marks
Differentiate y=(x+4)^{1/2} from first principle.
Solution
  1. Let y=\sqrt{x+4}. Give x a small increment \Delta x; then y changes to y+\Delta y.
    y+\Delta y=\sqrt{x+\Delta x+4}
  2. Subtract to get the change in y.
    \Delta y=\sqrt{x+\Delta x+4}-\sqrt{x+4}
  3. Divide by \Delta x and rationalise the numerator by multiplying with its conjugate.
    \dfrac{\Delta y}{\Delta x}=\dfrac{(x+\Delta x+4)-(x+4)}{\Delta x\left(\sqrt{x+\Delta x+4}+\sqrt{x+4}\right)}=\dfrac{1}{\sqrt{x+\Delta x+4}+\sqrt{x+4}}
  4. Take the limit as \Delta x\to0.
    \dfrac{dy}{dx}=\lim_{\Delta x\to0}\dfrac{1}{\sqrt{x+\Delta x+4}+\sqrt{x+4}}=\dfrac{1}{2\sqrt{x+4}}
\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x+4}}=\dfrac12(x+4)^{-1/2}
2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (iv), old syllabus4 marks
Differentiate y=\sin^{-1}\left(\dfrac{2x}{1+x^{2}}\right) w.r.t. x.
Solution
Method 1: substitution
  1. Put x=\tan\theta, so \theta=\tan^{-1}x. The inside becomes a double-angle formula.
    \dfrac{2x}{1+x^{2}}=\dfrac{2\tan\theta}{1+\tan^{2}\theta}=\sin2\theta
  2. For |x|<1, 2\theta lies in \left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right), so \sin^{-1}(\sin2\theta)=2\theta.
    y=2\theta=2\tan^{-1}x
  3. Differentiate.
    \dfrac{dy}{dx}=\dfrac{2}{1+x^{2}}
Method 2: chain rule (check)
  1. With u=\dfrac{2x}{1+x^{2}}: u'=\dfrac{2(1+x^{2})-2x(2x)}{(1+x^{2})^{2}}=\dfrac{2(1-x^{2})}{(1+x^{2})^{2}} and 1-u^{2}=\dfrac{(1-x^{2})^{2}}{(1+x^{2})^{2}}.
    \dfrac{dy}{dx}=\dfrac{u'}{\sqrt{1-u^{2}}}=\dfrac{2(1-x^{2})}{(1+x^{2})^{2}}\cdot\dfrac{1+x^{2}}{1-x^{2}}=\dfrac{2}{1+x^{2}}\quad(|x|<1)
\dfrac{dy}{dx}=\dfrac{2}{1+x^{2}}\quad(|x|<1)
2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (vii) OR, old syllabus4 marks
Evaluate \lim_{x\to0}\left(\dfrac{x\sin4x}{1-\cos2x}\right).
Solution
  1. Substitution gives \dfrac00. Use 1-\cos2x=2\sin^{2}x.
    \dfrac{x\sin4x}{1-\cos2x}=\dfrac{x\sin4x}{2\sin^{2}x}
  2. Arrange the factors so that standard limits \dfrac{\sin4x}{4x}\to1 and \dfrac{x}{\sin x}\to1 appear.
    =\dfrac12\cdot\dfrac{\sin4x}{4x}\cdot4x\cdot x\cdot\dfrac{1}{\sin^{2}x}=2\cdot\dfrac{\sin4x}{4x}\cdot\left(\dfrac{x}{\sin x}\right)^{2}
  3. Take the limit.
    \lim_{x\to0}\dfrac{x\sin4x}{1-\cos2x}=2\cdot1\cdot1^{2}=2
2
2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (ix), old syllabus4 marks
Differentiate implicitly to find \dfrac{dy}{dx} if x^{3}y^{2}+xy+y^{3}=10.
Solution
  1. Differentiate every term with respect to x. Use the product rule for x^{3}y^{2} and xy, and remember that each y term gains a factor \dfrac{dy}{dx}.
    \left(3x^{2}y^{2}+x^{3}\cdot2y\dfrac{dy}{dx}\right)+\left(y+x\dfrac{dy}{dx}\right)+3y^{2}\dfrac{dy}{dx}=0
  2. Collect the \dfrac{dy}{dx} terms on one side.
    \left(2x^{3}y+x+3y^{2}\right)\dfrac{dy}{dx}=-\left(3x^{2}y^{2}+y\right)
  3. Divide.
    \dfrac{dy}{dx}=-\dfrac{3x^{2}y^{2}+y}{2x^{3}y+x+3y^{2}}
\dfrac{dy}{dx}=-\dfrac{3x^{2}y^{2}+y}{2x^{3}y+x+3y^{2}}
2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (xii) OR, old syllabus4 marks
Find the second derivative of y=x^{2}e^{2x}.
Solution
  1. First derivative by the product rule.
    y'=x^{2}(2e^{2x})+e^{2x}(2x)=2e^{2x}(x^{2}+x)
  2. Differentiate again with the product rule on 2e^{2x} and x^{2}+x.
    y''=2\left[2e^{2x}(x^{2}+x)+e^{2x}(2x+1)\right]
  3. Take out 2e^{2x} and simplify.
    y''=2e^{2x}\left(2x^{2}+2x+2x+1\right)=2e^{2x}(2x^{2}+4x+1)
\dfrac{d^{2}y}{dx^{2}}=2e^{2x}(2x^{2}+4x+1)
2026 Annual I, Old syllabus (Curriculum 2006), Q.3, old syllabus8 marks
Test continuity of the function f(x)=\begin{cases}\dfrac{x^{2}-4}{x-2}, & x<2\\ \dfrac53x+\dfrac23, & 2\le x<5\\ x^{2}-16, & x\ge5\end{cases} at x=2 and x=5.
Solution

Condition for continuity

f is continuous at x=c if f(c) is defined, \lim_{x\to c}f(x) exists (the left-hand and right-hand limits are equal) and the limit equals f(c). At each point we take the value from the piece whose condition includes the point, and each one-sided limit from the piece on that side.

At x = 2

  1. Value from the middle piece (2\le x<5).
    f(2)=\dfrac53(2)+\dfrac23=\dfrac{10+2}{3}=4
  2. Left-hand limit from the first piece; cancel x-2.
    \lim_{x\to2^-}\dfrac{(x-2)(x+2)}{x-2}=\lim_{x\to2^-}(x+2)=4
  3. Right-hand limit from the middle piece.
    \lim_{x\to2^+}\left(\dfrac53x+\dfrac23\right)=4
  4. Both limits equal the value.
    \lim_{x\to2}f(x)=4=f(2)\ \Longrightarrow\ \text{continuous at }x=2

At x = 5

  1. Value from the last piece (x\ge5).
    f(5)=25-16=9
  2. Left-hand limit from the middle piece.
    \lim_{x\to5^-}\left(\dfrac53x+\dfrac23\right)=\dfrac{25+2}{3}=9
  3. Right-hand limit from the last piece.
    \lim_{x\to5^+}(x^{2}-16)=9
  4. All three agree.
    \lim_{x\to5}f(x)=9=f(5)\ \Longrightarrow\ \text{continuous at }x=5

Conclusion

f is continuous at x=2 and at x=5. In the graph the line y=x+2 (for x<2) meets the line y=\tfrac53x+\tfrac23 at (2,4), and that line meets the parabola y=x^{2}-16 at (5,9).

f\ \text{is continuous at }x=2\ \text{and at }x=5
2026 Annual I, Old syllabus (Curriculum 2006), Q.6 OR, old syllabus8 marks
Examine the extreme values of f(x)=\sin x\cos x, where x\in[0,\pi].
Solution

Simplify

Using \sin2x=2\sin x\cos x, the function is f(x)=\dfrac12\sin2x. It is continuous on the closed interval [0,\pi], so it has an absolute maximum and minimum there, found among the critical points and the endpoints.

Critical points

  1. Differentiate.
    f'(x)=\dfrac12\cdot2\cos2x=\cos2x
  2. Solve f'(x)=0 for x in [0,\pi], that is 2x in [0,2\pi].
    \cos2x=0\ \Longrightarrow\ 2x=\dfrac{\pi}{2},\ \dfrac{3\pi}{2}\ \Longrightarrow\ x=\dfrac{\pi}{4},\ \dfrac{3\pi}{4}

Second derivative test

  1. Differentiate again.
    f''(x)=-2\sin2x
  2. At x=\tfrac{\pi}{4}.
    f''\!\left(\tfrac{\pi}{4}\right)=-2\sin\tfrac{\pi}{2}=-2<0\ \Longrightarrow\ \text{maximum},\quad f\!\left(\tfrac{\pi}{4}\right)=\tfrac12
  3. At x=\tfrac{3\pi}{4}.
    f''\!\left(\tfrac{3\pi}{4}\right)=-2\sin\tfrac{3\pi}{2}=2>0\ \Longrightarrow\ \text{minimum},\quad f\!\left(\tfrac{3\pi}{4}\right)=-\tfrac12

Endpoints and conclusion

At the endpoints f(0)=0 and f(\pi)=0. Comparing 0, \tfrac12, -\tfrac12 and 0: the maximum value of f on [0,\pi] is \tfrac12, at x=\tfrac{\pi}{4}, and the minimum value is -\tfrac12, at x=\tfrac{3\pi}{4}. These relative extrema are also the absolute extrema. The graph is half of a sine wave squeezed to period \pi and height \tfrac12: it rises from 0 to its peak at \tfrac{\pi}{4}, falls through 0 at \tfrac{\pi}{2} to its lowest point at \tfrac{3\pi}{4}, and returns to 0 at \pi.

\max=\tfrac12\ \text{at }x=\tfrac{\pi}{4},\qquad\min=-\tfrac12\ \text{at }x=\tfrac{3\pi}{4}

Exam focus

  1. Find two constants so that a three-piece function is continuous at two joining points; or test such a function for continuity at x=2 and x=5. (long)
    Asked as the 8-mark Q.3 in every 2026 paper (Sets B and D, and both old-syllabus papers).
    Go to answer
  2. Find the critical points of a trigonometric function on [0,\pi], classify them by the second derivative test, and find the absolute extrema. (long)
    2026 Sets B and D Q.4 (8 marks) and the Curriculum 2006 Q.6 OR; endpoints must be compared too.
    Go to answer
  3. Find the intervals on which a cubic function is increasing or decreasing. (problem)
    2026 Sets B and D Q.2 (iii) OR; the method is a short sign table of the factorised derivative.
    Go to answer
  4. Use differentials to find the approximate (percentage) error in the volume of a cube or sphere, or the approximate value of a root or trigonometric ratio. (numerical)
    2026 Sets B and D Q.2 (viii), MCQs 19 and 20, and sin 46 degrees in the Curriculum 2000 paper.
    Go to answer
  5. Differentiate a quotient or a product of an exponential and a logarithm, and simplify. (problem)
    2026 Sets B and D Q.2 (ii) and Q.2 (xi) OR; simplification earns the last mark.
    Go to answer
  6. Differentiate from first principles: x^{n}, \sqrt x, \sqrt{x+a}, \dfrac1x, \sin x and \cos x. (long)
    Standard derivations; the Curriculum 2006 paper asked for (x+4)^{1/2} from first principles.
    Go to answer
  7. Evaluate limits with \dfrac00 forms, the standard limit \dfrac{\sin x}{x}\to1 and the e-type limits. (short)
    All of Exercise 2.1, and the 2026 old-syllabus papers (Q.2 (vi) OR and Q.2 (vii) OR).
    Go to answer
  8. Derive and use the derivatives of the inverse trigonometric functions. (problem)
    The Curriculum 2000 paper set an 8-mark question on them; the derivation of \sin^{-1}x and \tan^{-1}x is a standard short question.
    Go to answer
  9. Find dy/dx by implicit and by parametric differentiation. (problem)
    Asked in both 2026 old-syllabus papers (Q.2 (ix) and Q.2 (xi) OR) and in Exercise 2.7.
    Go to answer
  10. Determine where a function is discontinuous, and whether it is continuous on a given interval. (problem)
    The core skill of Exercise 2.2; MCQs on continuity and differentiability are common.
    Go to answer

Mark this unit as done to fill the Class 12 Mathematics ring.

Next, unit 3
Integration

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