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C H A P T E R

22Differentiation

Techniques

ObjectivesTo differentiate functions having negative integer powers.

To understand and use the chain rule.

To differentiate rational powers.

To find second derivatives of functions.

To use limits to define continuity at a point.

To use differentiation techniques to sketch graphs of functions.

To apply differentiation techniques to solving problems.

The work in this chapter is not part of the content for VCE Mathematical Methods (CAS)

Units 1 & 2. It is included as a useful foundation for VCE Mathematical Methods (CAS)

Units 3 & 4.

22.1 Differentiating xn where n isa negative integerIn earlier chapters the differentiation of polynomial functions has been introduced and

applications of differentiation of such functions have been considered. In this section we add to

the family of functions for which we can find the derived functions. In particular, we will

consider functions which involve linear combinations of powers of x where a power may be a

negative integer.

e.g. f : R \ {0} → R, f (x) = x−1

f : R \ {0} → R, f (x) = 2x + x−1

f : R \ {0} → R, f (x) = x + 3 + x−2

Note: We have reintroduced the function notation which emphasises the need for consideration

of domain.

582

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Chapter 22 — Differentiation Techniques 583

Example 1

Let f : R \ {0} → R, f (x) = 1

x. Find f ′(x) by first principles.

Solution

y

x

x, P 1x

Q x + h, ––––1x + h

0

Gradient of chord PQ = f (x + h) − f (x)

x + h − x

=1

x + h− 1

xh

= x − (x + h)

(x + h)x× 1

h

= −h

(x + h)x× 1

h

= −1

(x + h)x

The gradient of the curve at P = limh→0

−1

(x + h)x= −1

x2= x−2, i.e. f ′(x) = −x−2

Example 2

Let f : R \ {0} → R, f (x) = x−3. Find f ′(x) by first principles.

Solution

y

x

(x, x–3)

(x + h, (x + h)–3)

0

Q

P

Gradient of chord PQ = (x + h)−3 − x−3

h

= x3 − (x + h)3

(x + h)3x3× 1

h

= x3 − (x3 + 3x2h + 3xh2 + h3)

(x + h)3x3× 1

h

= −3x2h − 3xh2 − h3

(x + h)3x3× 1

h

= −3x2 − 3xh − h2

(x + h)3x3

The gradient of the curve at P = limh→0

−3x2 − 3xh − h2

(x + h)3x3

= −3x2

x6

= −3

x4

= −3x−4

i.e. f ′(x) = −3x−4

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584 Essential Mathematical Methods 1 & 2 CAS

We are now in a position to state the generalisation of the result we found in Chapter 19.

For f (x) = xn , f ′(x) = nxn−1, n is a non-zero integer.

For f (x) = c, f ′(x) = 0, where c is a constant.

We note that for n ≤ −1 we take the domain of f to be R \ {0}, and for n ≥ 1 we take

the domain of f to be R.

Example 3

Find the derivative of x4 − 2x−3 + x−1 + 2, x �= 0.

Solution

If f (x) = x4 − 2x−3 + x−1 + 2, x �= 0

f ′(x) = 4x3 − 2(−3x−4) + (−x−2) + 2(0)

= 4x3 + 6x−4 − x−2, x �= 0

Example 4

Find the derivative f ′ of f : R \ {0} → R, f (x) = 3x2 − 6x−2 + 1.

Solution

f ′: R \ {0} → R, f ′(x) = 3(2x) − 6(−2x−3) + 1(0)

= 6x + 12x−3

Example 5

Find the gradient of the curve determined by the function

f : R \ {0}, f (x) = x2 + 1

xat the point (1, 2).

Solution

f ′: R \ {0} → R, f ′(x) = 2x + (−x−2)

= 2x − x−2

and f ′(1) = 2 − 1

= 1

The gradient of the curve is 1 at the point (1, 2).

Example 6

Show that the gradient of the function f : R \ {0} → R, f (x) = x−3 is always negative.

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Solution

f ′: R \ {0} → R, f ′(x) = −3x−4

= −3

x4

x−4 is positive for all x and thus f ′(x) < 0 for all x �= 0.

Exercise 22A

1 Differentiate each of the following with respect to x :Examples 3, 4

a 3x−2 + 5x−1 + 6 b3

x2+ 5x2 c

5

x3+ 4

x2+ 1

d 3x2 + 5

3x−4 + 2 e 6x−2 + 3x f

3x2 + 2

x

2 Find the derivative of each of the following:

a3z2 + 2z + 4

z2, z �= 0 b

3 + z

z3, z �= 0 c

2z2 + 3z

4z, z �= 0

d 9z2 + 4z + 6z−3, z �= 0 e 9 − z−2, z �= 0 f5z − 3z2

5z, z �= 0

3 a Carefully sketch the graph of f (x) = 1

x2, x �= 0.Examples 1, 2

b Let P be the point (1, f (1)) and Q the point (1 + h, f (1 + h)). Find the gradient of the

chord PQ.

c Hence find the gradient of the curve f (x) = 1

x2at x = 1.

d Find the equation of the normal to the curve with equation y = x−2 at the point (1, 1).

4 Find the gradient of each of the following curves at the given point:Example 5

a y = x−2 + x3, x �= 0, at(2, 8 1

4

)b y = x − 2

x, n �= 0, at

(4, 1

2

)

c y = x−2 − 1

x, x �= 0, at (1, 0) d y = x(x−1 + x2 − x−3), x �= 0, at (1, 1).

5 Find the x-coordinate of the point on the curve with equation f (x) = x−2 and gradient:

a 16 b −16

22.2 The chain ruleAn expression such as f (x) = (x3 + 1)2 may be differentiated by expanding and then

differentiating each term. This method is a great deal more tiresome for an expression

such as f (x) = (x3 + 1)30. We transform f (x) = (x3 + 1)2 into two simpler functions

defined by:

h(x) = x3 + 1(= u) and g(u) = u2(= y)

which are ‘chained’ together: xh−→u

g−→y

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We may exploit this connection to differentiate. We use Leibniz notation to explore

this idea.

Consider f (x) = (x3 + 1)2 at the point where x = 2.

u y

u

Q

R

S

u = x3 + 1

y = u2

x 00 91 2

δx

δyδu

δu

P

80

60

40

20(0, 1)

–2 –1 0 1 2

y

x

Z

W

δy

δx

y = (x3 + 1)2

When x = 2, u = 9 When u = 9, y = 81 When x = 2, y = 81

P is the point (2, 9) R is the point (9, 81) Z is the point (2, 81)

A section of a spreadsheet (below) illustrates the connection between the gradients of the

chords PQ, RS and WZ. It can be seen that as �x gets smaller so does �u and�y

�x= �y

�u× �u

�x.

By considering �u becoming smaller and hence �x becoming smaller, it is seen that

dy

dx= dy

du· du

dx

This is called the chain rule for differentiation.

x u y �x �u �y�u

�x

�y

�u

�u

�x× �y

�u

�y

�x

1.60000 5.09600 25.96922 0.40000 3.90400 55.03078 9.76000 14.09600 137.57696 137.57696

1.80000 6.83200 46.67622 0.20000 2.16800 34.32378 10.84000 15.83200 171.61888 171.61888

1.90000 7.85900 61.76388 0.10000 1.14100 19.23612 11.41000 16.85900 192.36119 192.36119

1.99000 8.88060 78.86504 0.01000 0.11940 2.13496 11.94010 17.88060 213.49614 213.49614

1.99900 8.98801 80.78425 0.00100 0.01199 0.21575 11.99400 17.98801 215.74816 215.74816

1.99990 8.99880 80.97840 0.00010 0.00120 0.02160 11.99940 17.99880 215.97480 215.97480

1.99999 8.99988 80.99784 0.00001 0.00012 0.00216 11.99994 17.99988 215.99748 215.99748

From the spreadsheet it can be seen that the gradient of u = x3 + 1 at x = 2 is 12, and the

gradient of y = u2 at u = 9 is 18. The gradient of y = (x3 + 1)2 at x = 2 is 216. The chain

rule is used to confirm this.

du

dx= 3x2 and at x = 2

du

dx= 12

dy

du= 2u and at u = 9

dy

du= 18

dy

dx= dy

du· du

dx= 216

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Example 7

Find the derivative of y = (3x+ 4)20.

Solution

Let u = 3x + 4 then y = u20

Sodu

dx= 3 and

dy

du= 20u19

∴ dy

dx= dy

du· du

dx

= 20u19 · 3

= 60(3x + 4)19

Example 8

Find the gradient of the curve with equation y = 16

3x2 + 1at the point (1, 4).

Solution

Let u = 3x2 + 1 then y = 16u−1

Sodu

dx= 6x and

dy

du= −16u−2

∴ dy

dx= dy

du· du

dx

= −16u−2 · 6x

= −96x

(3x2 + 1)2

∴ at x = 1 the gradient is−96

16= −6.

Example 9

Differentiate y = (4x3 − 5x)−2.

Solution

Let u = 4x3 − 5x

then y = u−2

anddu

dx= 12x2 − 5 and

dy

du= −2u−3

∴ dy

dx= dy

du· du

dx

= −(2u−3) · (12x2 − 5)

= −2(12x2 − 5)

(4x3 − 5x)3

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Example 10

Use the chain rule to provedy

dx= nxn−1 for y = xn and n a negative integer.

(Assume the result for n a positive integer and x > 0.)

Solution

Let n be a negative integer and y = xn. Then y = 1

x−nas −n is a positive integer.

Let u = x−n then y = 1

u= u−1

Thusdu

dx= −nx−n−1 (−n is a positive integer) and

dy

du= −u−2

∴ dy

dx= dy

du· du

dx

= −u−2 · (−nx−n−1)

= nx−n−1(x−n)−2

= nx−n−1(x2n)

= nxn−1

With function notation the chain rule is stated as:

( f ◦ g)′(x) = f ′(g(x))g′(x)

where f ◦ g(x) = f (g(x)).

Example 11

Given that f (x) = (x2 + 1)3, find f ′(x).

Solution

Now f = k ◦ g where k(x) = x3 and g(x) = x2+ 1

It follows that k′(x) = 3x2 and g′(x) = 2x

By the chain rule f ′(x) = k ′(g(x))g ′(x)

We have in this case f ′(x) = 3(g(x))2 2x

which yields f ′(x) = 6x(x2 + 1)2

Exercise 22B

1 Differentiate each of the following with respect to x :Examples 7, 8, 9

a (x − 1)30 b (x5 − x10)20 c (x − x3 − x5)4

d (x2 + 2x + 1)4 e (x2 + 2x)−2, x �= −2, 0 f

(x2 − 2

x

)−3

, x �= 0

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2 a Find the derivative of f : R → R, f (x) = (2x3 + 1)4.

b Find the gradient of f : R → R, f (x) = (2x3 + 1)4 at the point (1, 81).

3 a Find the gradient of the curve with equation y = 1

x + 3at the point

(1, 1

4

).

b Find the gradient of the curve with equation y = 1

(x + 3)3, at the point

(1, 1

64

).

4 The diagram is a sketch graph

of a function with f (x) = 1

2x + 3. x = – –3

2

y

x0

0, –13

a Find the gradient of the

curve at the points(0, 1

3

).

b Find the coordinates of the

points on the curve for which

the gradient is −2

9.

5 The diagram is a sketch graph of

the functions y = 1

xand y = −1

x.

y

x

y = – 1x

y = – – 1x

y = – 1x

P(–1, 1)

Q(–1, –1)

2

1

0 1 2 3 4–1–1

–2

–2–3–4

–a Find the gradient of y = 1

xat the point

(2, 1

2

).

b Without further calculation

state the gradient of y = −1

xat

(2, −1

2

).

c Find the equation of the tangent at the point (1, 1) of y = 1

x.

d Find the equation of the tangent at the point (1, –1) of y = −1

x.

e Find the equations of the tangents at points P and Q and find their point of intersection.

f Draw sketch graphs of y = 1

xand y = −1

xon the same set of axes and draw in the four

tangents.

22.3 Differentiating rational powers(x

p/q)

Using the chain rule in the formdy

du= dy

dx· dx

du

with y = u 1 = dy

dx· dx

dy

and thusdy

dx= 1

dx

dy

, fordx

dy�= 0

Let y = x1n , where n ∈ Z \ {0} and x > 0.

Then yn = x anddx

dy= nyn−1.

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From the above resultsdy

dx= 1

nyn−1

= 1

n

(x

1n

)n−1

= 1

nx

1n −1

For y = x1n ,

dy

dx= 1

nx

1n −1; n ∈ Z \ {0} and x > 0.

This result may now be extended to any rational power.

Let y = xpq , where p, q ∈ Z \ {0}.

Write y =(

x1q

)p

. Let u = x1q . Then y = up.

The chain rule yieldsdy

dx= dy

du· du

dx

= pu p−1 · 1

qx

1q −1

= p

(x

1q

)p−1

· 1

qx

1q −1

= p

qx

pq − 1

q x1q −1

= p

qx

pq −1

Thus the result for any non-zero rational power has been established and, in fact, it is true for

any non-zero real power.

For f (x) = xa, f ′(x) = axa−1, for x > 0 and a ∈ R.

These results have been stated for

x > 0, as (−3)12 is not defined

although (−2)13 is.

The graphs of y = x12 , y = x

13 and

y = x14 are shown.

The domain of each has been taken

to be R +.

1.5

1.0

0.5

–0.5

–0.5

0.5 1.0 1.5 2.0 2.5 3.00

y

x

y = x

y = x

y = x14

13

12

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The figure to the right is the graph of the function f : R → R, f (x) = x13 .

Note that the values shown here

are −0.08 ≤ x ≤ 0.08.

From this it can be seen that the

tangent at the origin of y = x13

is on the y-axis.

–0.2

0.4

0.2

–0.4

0.02–0.02–0.04–0.06–0.08 0.04 0.06 0.080

y

x

3y = x

1

Use a calculator to investigate

graphs of this type further.

Example 12

Find the derivative of each of the following with respect to x :

a 4x23 b x

15 − 2x−3

Solution

a Let y = 4x23 , then

dy

dx= 4 × 2

3× x

23 −1 = 8

3x

−13

b Let y = x15 − 2x−3, then

dy

dx= 1

5x

15 −1 − 2 × −3x−3−1

= 1

5x

−45 + 6x−4

Exercise 22C

1 Find the derivative of each of the following with respect to x :Example 12

a x13 b x

32 , x > 0 c x

52 − x

32 , x > 0

d 2x12 − 3x

53 , x > 0 e x− 5

6 , x > 0 f x− 12 − 4, x > 0

2 Find the derivative of each of the following with respect to x :

a√

1 + x2 b 3√x + x2 c (1 + x2)−12 d (1 + x)

13

3 a Find the gradient of y = x13 at each of the following points:

i

(1

8,

1

2

)ii

(−1

8, −1

2

)iii (1, 1) iv (−1, −1)

b Comment on your results.

4 Consider the graphs of y = x12 and y = x

13 for x > 0.

a Find

{x : x

12 < x

13

}.

b Find the values for x for which the gradient of y = x12 is greater than the gradient

of y = x13 .

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5 Differentiate each of the following with respect to x :

a(2 − 5

√x)2

b(3√

x + 2)2

c2 + √

x

x2

dx2 + 2√

xe 3

√x

(x2 + 2

)

22.4 The second derivativeFor f : R →R, f (x) = 2x2 + 4x + 1, the derived function has rule f ′(x) = 4x + 4,

i.e. the derivative of 2x2 + 4x + 1 is 4x + 4.

The second derivative of 2x2+ 4x + 1 is 4, i.e. the derivative of 4x + 4.

The function notation for the second derivative is f ′′(x) = 4.

In the Leibniz notation

if y = 2x2 + 4x + 1,dy

dx= 4x + 4 and the second derivative is written

d2 y

dx2, and

d2 y

dx2= 4.

Example 13

For each of the following find f ′′(x).

a f (x) = 3x3 + 2x−1 + 1, x �= 0 b f (x) = x12 + 3x−4 + 1, x > 0

c f (x) = x4 + x− 32 + 1, x > 0

Solution

a f ′(x) = 9x2 − 2x−2, x �= 0

f ′′(x) = 18x+ 4x−3, x �= 0

b f ′(x) = 1

2x− 1

2 − 12x−5, x > 0

f ′′(x) = −1

4x− 3

2 + 60x−6, x > 0

c f ′(x) = 4x3 − 3

2x− 5

2 , x > 0

f ′′(x) = 12x2 + 15

4x− 7

2 , x > 0

If S(t) denotes the displacement of a body at time t,

then S′(t) gives the velocity at time t,

and S′′(t) gives acceleration at time t.

Exercise 22D

1 Find f ′′(x) for each of the following:Example 13

a f (x) = x3 + 2x + 1 b f (x) = 3x + 2 c f (x) = (3x + 1)4

d f (x) = x12 + 3x3, x > 0 e f (x) = (x6 + 1)3 f f (x) = 5x2 + 6x−1 + 3x

32

2 For each of the following findd2 y

dx2:Example 13

a y = 3x3 + 4x + 1 b y = 6 c y = 6x2 + 3x + 1

d y = (6x + 1)4 e y = (5x + 2)4 f y = x3 + 2x2 + 3x−1

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3 The height of a stone is (20t – 4.9t2) metres at a time t seconds after it is thrown. What is

the acceleration?

4 After t seconds, the x-coordinate of a particle moving along the x-axis is given by

x(t) = 4t – 3t3, where the units on the x-axis are in metres.

a Find:

i the x-coordinate after 2 seconds of motion

ii the velocity of the particle at the start

iii the velocity of the particle after half a second

iv the velocity of the particle after 2 seconds.

b When is the acceleration zero?

c What is the average velocity during the first 2 seconds?

22.5 Sketch graphsIn Chapter 20 it was demonstrated how calculus may help us sketch the graphs of polynomial

functions. In this section we show how the same techniques may be applied to non-polynomial

functions.

For example: Let f : R \ {0} → R, f (x) = 1

x2+ x .

This function is not defined at x = 0.

1 Consider the behaviour of the function as the magnitude of x becomes large.

x f (x)

1 2.000000000000000

2 2.250000000000000

4 4.062500000000000

8 8.015625000000000

16 16.003906250000000

32 32.000976562500000

64 64.000244140625000

128 128.000061035156250

256 256.000015258789063

512 512.000003814697266

1024 1024.00000095367432

2048 2048.00000023841858

This section of a spreadsheet demonstrates the

behaviour of increasing x . This has been done

by using increasing powers of 2.

We see that as x → ∞, f (x) → x .

This shows us that the line y = x is an

oblique asymptote for the graph.

It is clear also that as x → −∞, f (x) → x .

The behaviour of the function as the magnitude of x approaches 0 is demonstrated below.

A spreadsheet has been used and the values of x considered are 20, 2−1, 2−2, 2−3, etc.

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a

x f (x)

1.0000000000 2.000000000000

0.5000000000 4.500000000000

0.2500000000 16.250000000000

0.1250000000 64.125000000000

0.0625000000 256.062500000000

0.0312500000 1024.031250000000

0.0156250000 4096.015625000000

0.0078125000 16384.007812500000

0.0039062500 65536.003906250000

Spreadsheet a shows the behaviour as x

approaches 0 from the right.

f (x) becomes increasingly large.

We write limx→0+

f (x) = ∞.

b

x f (x)

–1.0000000000 0.000000000000

–0.5000000000 3.500000000000

–0.2500000000 15.750000000000

–0.1250000000 63.875000000000

–0.0625000000 255.937500000000

–0.0312500000 1023.968750000000

–0.0156250000 4095.984375000000

–0.0078125000 16383.992187500000

–0.0039062500 65535.996093750000

Spreadsheet b shows the behaviour as x

approaches 0 from the left.

Again f (x) becomes increasingly large.

We write limx→0−

f (x) = ∞.

2 Consider the axis intercepts.

The function cannot have a y-axis intercept, as it is not defined at x = 0. It would cross the

x-axis when1

x2+ x = 0, i.e. when x3 = −1, which implies x = −1.

3 Consider points of zero gradient (turning points).

For f (x) = 1

x2+ x we have f ′(x) = −2

x3+ 1 and f ′(x) = 0 implies x = 2

13 , i.e. there is a

point on the curve at x = 213 at which the gradient is zero.

As the graph is not continuous for R, it is advisable to use a gradient chart for x > 0 only.

f ′(x)

213

shape of f

– +0

x

6

4

2

0 2 4 6–2

–2

–4

–4–6

(1.26, 1.89)

y

x

y = –– + x1x2

Thus there is a minimum at the point (1.26, 1.89)

(coordinates correct to 2 decimal places).

The spreadsheet above shows how rapidly the

graph of y = 1

x2+ x moves towards y = x.

At x = 4, f (x) = 4.0625 and at x = 8,

f (x) = 8.015625.

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Example 14

For f : R \ {−1} → R, f (x) = x2 + 3

x + 1, sketch the graph.

Solution

By division, f (x) = x − 1 + 4

x + 1.

Behaviour as magnitude of x increases

As x → ∞, f (x) → x – 1, from above.

As x → –∞, f (x) → x – 1, from below.

There is an oblique asymptote with equation y = x − 1.

Behaviour as x→−1

limx→−1+

f (x) = ∞ and limx→−1−

f (x) = −∞

There is a vertical asymptote x = –1.

Axis intercepts

When x = 0, f (x) = 3. There is no x-axis intercept as x2 + 3 �= 0 for all x ∈ R.

Turning points

For f (x) = x − 1 + 4

x + 1

f ′(x) = 1 − 4

(x + 1)2

Where f ′(x) = 0

1 − 4

(x + 1)2= 0

which implies (x + 1)2 = 4

i.e. x + 1 = 2 or x + 1 = –2

∴ x = 1 or x = –3

f (1) = 2 and f (–3) = –6

We consider two gradient charts.

First for x > –1.

1

f ′(x)

shape of f

– +0

x

∴ a minimum at (1, 2)

Next, for x < −1.

+

–3

f ′(x)

shape of f

–0

x

∴ a maximum at (–3, –6)

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We are now in a position to sketch the graph.

y

x

x = –1

y = x – 1

(0, 3)

(–3, –6)

10

0 5 10 15

–5

–10

–15

–15 –10 –5

5

y = ––––––x2 + 3x + 1

Exercise 22E

1 The equation of a curve is y = 4x + 1

x.

a Find the coordinates of the turning points.

b Find the equation of the tangent to the curve at the point where x = 2.

2 Find the x-coordinates of the points on the curve y = x2 − 1

xat which the gradient of the

curve is 5.

3 Find the gradient of the curve y = 2x − 4

x2at the point where the curve crosses the x-axis.

4 For the curve y = x − 5 + 4

x, find:

a the coordinates of the points of intersection with the axes

b the equations of all asymptotes c the coordinates of all turning points.

Use this information to sketch the curve.

5 If x is positive, find the least value of x + 4

x2.

6 For positive values of x, sketch the graph of y = x + 4

xand find the least value of y.

7 Sketch the graphs of each of the following, indicating the coordinates of the axes interceptsExample 14

and the turning point, and the equations of asymptotes.

a y = x + 1

x, x �= 0 b y = 1

x2− x, x �= 0

c y = x + 1 + 1

x + 3, x �= 3 d y = x3 + 243

x, x �= 0

e y = x − 5 + 1

x, x �= 0 f y = x2 − 4

x + 2, x �= −2

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Review


Chapter summary

The general result for differentiating functions, including powers of x with negative

integers:

For f (x) = xn, f ′(x) = nxn−1, n a non-zero integer. For f (x) = 1, f ′(x) = 0.

We note that for n ≤ −1, we take the domain of f to be R \ {0}, and for n ≥ 1 we take the

domain of f to be R.

The chain rule is often used to differentiate some more complicated functions by

transforming the original function into two simpler functions:

e.g. f (x) is transformed to h(x) and g(u), which are ‘chained’ together as

xh−→u

g−→y

Using Leibniz notation the chain rule is stated asdy

dx= dy

du· du

dx.

With function notation the chain rule is stated as

( f ◦ g)′(x) = f ′(g(x))g′(x) where ( f ◦ g)(x) = f (g(x))

The general result for any non-zero real power:

For f (x) = xa, f ′(x) = axa − 1, for x > 0 and a ∈ R.

The function for the second derivative is f ′′.

In Leibniz notation it is written asd 2 y

dx2.

When using calculus as an aid to sketching graphs of polynomial and non-polynomial

functions, the following should be considered:

1 The behaviour of the function as the magnitude of x becomes large

2 The axis intercepts

3 The points of zero gradient (turning points).

Multiple-choice questions

1 If f (x) = 4x4 − 12x2

3xthen f ′(x) equals

A16x3 − 24x

3B 4x2 − 4 C

16x3 − 24x

3x

D 4x2 − 8x E8x3 − 16x

3x

2 If f (x) = 2xpq , where p and q are integers, f ′(x) equals

A 2x(p−q)

q B 2pxpq −1

C 2 D2p

qx

(p−q)q E

2p

qx

3 For f : R \ {2} → R, where f (x) = 4 + 4

2 − x, f ′(x) > 0 for

A R \ {2} B R C x < 2 D x > 2 E x > 4

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Rev

iew


4 A particle moves in a straight line so that its position x cm from a fixed point O at time

t s (t ≥ 0) is given by x = −t3 + 7t2 − 14t + 6. The particle’s acceleration at t = 3 is

A −4 cm/s2 B 3 cm/s2 C 4 cm/s2 D 8 cm/s2 E 0 cm/s2

5 Let y = f (g(x)), where g(x) = x3, thendy

dxequals

A 3x2 f ′(x3) B 3x2 f (x3) C 2x f (x) f ′(x3)

D 2 f (x) f ′(x3) E 3x2

6 The graph defined by the rule f (x) = x + 1

xhas a local minimum at (a, f (a)).

The value of a is

A −1 B 2 C −5

2D

5

2E 1

7 Which of the following is not true for the curve of y = f (x), where f (x) = x15 ?

A The gradient is defined for all real numbers.

B The curve passes through the origin.

C The curve passes through the points with coordinates (1, 1) and (−1,−1).

D For x > 0 the gradient is positive.

E For x > 0 the gradient is decreasing.

8 Which of the following is not true for y = f (x), where f (x) = x34 ?

A The maximal domain of the function is R+ ∪ {0}.

B f (x) > x for all x > 1.

C The curve of y = f (x) passes through the points with coordinates (1, 1).

D For x > 0 the gradient of the curve is positive.

E For x > 0 the gradient of the curve is decreasing.

9 The derivative of (5x2 + 2x)n is

A n(10x + 2)(5x2 + 2x)n−1 B (5x2 + 2x)n−1 C (10x + 2)n

D n(5x2 + 2x)n−1 E 10x2n−1 + 2xn−1

10 The graph of the function with rule y = k

2(x2 + 1)has gradient 1 when x = 1.

The value of k is

A 1 B −1 C 4 D −4 E −1

4

Short-answer questions (technology-free)


a x−4 b 2x−3 c − 1

3x2d − 1

x4

e3

x5f

x2 + x3

x4g

3x2 + 2x

x2h 5x2 − 2

x


a x12 b 3√x c − 2

x13

d x43 e x− 1

3 f x− 13 + 2x

35

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Review


3 Differentiate each of the following with respect to x :

a (2x + 3)2 b 2(3x + 4)4 c (3 − 2x)−12 d

1

3 + 2x

e1

(2x − 1)23

f3√

2 + x2g

(2x2 − 3

x2

) 13

4 Find the gradient of each of the following curves at the given point:

a y = √x ; (9, 3) b y = 1

2x + 1; (0, 1) c y = 2

x2;

(4,

1

8

)

d y = 3 + 2

x; (1, 5) e y = √

x + 1; (8, 3) f y = (x2 − 7x − 8)3; (8, 0)

5 Find the coordinates of the point(s) on the curve with equation y = 1

xfor which the

gradient is −4.

6 Find the coordinates of the point(s) on the curve with equation y = √x for which the

gradient is 2.

Extended-response questions

1 A solid circular cylinder has radius r cm and height h cm. It has a fixed volume of 400 cm3.

a Find h in terms of r.

b Show that the total surface area, A cm2, of the cylinder is given by A = 2�r2 + 800

r.

c FinddA

dr.

d Solve the equationdA

dr= 0 for r.

e Find correct to 3 significant figures the minimum surface area of the cylinder.

f Sketch the graph of A against r.

2 A rectangle has sides of length x cm and y cm and the area of the rectangle is 16 cm2.

a Find y in terms of x .

b Show that the perimeter, P cm, is given by P = 2x + 32

x.

c Find the value of x for which the value of P is a minimum and find this value of P.

d Sketch the graph of P against x for x > 0.

3 The area of rectangle OABC is 120 cm2. Let the length of OC be x cm, CZ = 5 cm and

AX = 7 cm.

A

O C

X Y

Z

Ba Find the length of OA in terms of x .

b Find the length of OX in terms of x .

c Find the length of OZ in terms of x .

d Find the area, A cm2, of rectangle OXYZ in terms of x .

e Find the value of x for which the area, A cm2, is a minimum.

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Rev

iew


4 The curve with equation y = √x + 2 meets the x-axis at A and the y-axis at B.

a Find the coordinates of A and B.

b By using the chain rule finddy

dx.

c i Find the gradient of the curve where x = −1.

ii Find the equation of the tangent at the point where x = −1.

iii If the tangent meets the x-axis at C and the y-axis at D, find the distance CD.

d Find the values of x for whichdy

dx< 1.

5 An open rectangular box of height h cm has a horizontal rectangular base with side lengths

x cm and 2x cm. The volume of the box is 36 cm3:

a Express h in terms of x .

b Show that the total surface area of the box is given by A = 2x2 + 108

x.

c Calculate the values of x and h which make the total surface area a minimum.

d Sketch the graph of A against x for x > 0.

6 The prism shown in the diagram has a triangular cross-section. The ‘ends’ of the prism

shown are congruent right-angled triangles with the right angles at C and Z.

A

B

C

X

Y

Z

AX = CZ = BY = y cm, AC = XZ = 3x cm and

CB = ZY = 4x cm

The volume of the prism is 1500 cm3.

a Express y in terms of x .

b Show that the total surface area, S cm2, is

given by S = 12x2 + 3000

x.

c FinddS

dx.

d Find the minimum value of S.

CHAPTER 22mathsbooks.net/Cambridge 11 Methods/ch22.pdf · 2015-08-04 · P1: FXS/ABE P2: FXS...

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