HIGHERSCHOOLCERTIFICATEYEAR11ADVANCEDMATHEMATICS

CALCULUSFREESAMPLE

1

First Published by John Kinny-Lewis in 2018 John Kinny-Lewis 2018 National Library of Australia Cataloguing-in-publication data ISBN: 978-0-6484118-7-1 This book is copyright. Apart from any fair dealing for purposes of private study, research, criticism or review as permitted under the Copyright Act 1968, no part may be reproduced, stored in a retrieval system, or transmitted, in any form by any means, electronic, mechanical, photocopying, recording, or otherwise without prior written permission. Enquiries to be made to John Kinny-Lewis. Copying for educational purposes. Where copies of part or the whole of the book are made under Section 53B or Section 53D of the Copyright Act 1968, the law requires that records of such copying be kept. In such cases the copyright owner is entitled to claim payment. Typeset by John Kinny-Lewis Edited by John Kinny-Lewis

2

CONTENTS SET 1 5 SET 1 ANSWERS 17 SET 1 SOLUTIONS 18 SET 2 30 SET 2 ANSWERS 42 SET 2 SOLUTIONS 43 SET 3 55 SET 3 ANSWERS 67 SET 3 SOLUTIONS 68 SET 4 80 SET 4 ANSWERS 92 SET 4 SOLUTIONS 93

3

PREFACE This book provides a thorough revision of calculus for the new Year 11 Advanced Mathematics HSC syllabus. The book is divided into four sets. Each set contains four topic areas.

• Gradients of tangents provides questions on: continuous and discontinuous functions, gradients of secants and its relationship to the tangent at a point.

• Difference quotients provides questions on: the behaviour of a

function and its tangent at a point, the difference quotient

f (x +h)− f (x )h

, distance-time and velocity-time graphs.

• The derivative function and its graph provides questions on:

the concept of a limit, defining the derivative f '(x ) from first principles, sketching the derivative function for a given graph of a function, establishing that f '(x ) = 0 yields a stationary point, examining variable rates of change of non-linear functions.

• Calculating with derivatives provides questions on:

the use of the formula

ddx

(xn ) = nxn−1 , differentiating a constant

multiple of a function and the sum and difference of two functions, understanding and using the product, quotient and chain rules.

Each topic area contains questions with space for the student’s working. Answers and detailed solutions are provided at the end of each set. Harder questions are marked with #.

4

SET 1 Gradients of tangents Question 1

Sketch the function f (x ) = x 2 − 4 in the domain − 3 ≤ x ≤ 3.State if this function is continousor discontinous, giving reasons.

Question 2

Sketch the function f (x ) = x 2 − 4x − 2

in the domain − 3 ≤ x ≤ 3.State if this function is continousor discontinous, giving reasons.

Question 3

Sketch the function f (x ) = 12x

in the domain − 3 ≤ x ≤ 3.State if this function is continousor discontinous, giving reasons.

5

SET 1 Gradients of tangents Question 4

The secant intersects the parabola y = x 2

at 1,1( ) and 1+h,(1+h)2( ).Find the gradient of the secant.

Question 5

The secant intersect the parabola y = x 2

at 1,1( ) and 1.5,2.25( ).Find the gradient of the secant.

Question 6

The tangent to y = x 2 touches the curve at (1,1) and intersects the

x axis at 12

,0⎛⎝⎜

⎞⎠⎟as shown in the

adjacent diagram.Find the gradient of the tangent.

6

SET 1 Difference quotients Question 1

The tangent to the parabola y = 4x − x 2

touches the parabola at (1,3) and intersects the x axis at (-0.5,0).Find the gradient of the tangent andhence, or otherwise, determine if theparabola is increasing or decreasing.

Question 2

The tangent to the curve y = xtouches the parabola at (4,2) and intersects the y axis at (0,1).Find the gradient of the tangent anddetermine if the curve is:A) increasing at a decreasing rate.B) increasing at an increasing rate.C) decreasing at a decreasing rate.D) decreasing at an increasing rate.

Question 3

P x, f (x )( ) and Q x +h, f (x +h)( ) are points on the the curve f (x ) = x 2 + x.

Show that the gradient of the secant PQ is 2x +1+h.

7

SET 1Difference quotientsQuestion 4

A(1,5) and B(3,9) are points on the curve x = 6t − t2, as shown in the adjacent diagram.Find the average velocity from A to B.

Question 5

A particle moves in a straight line with a displacement from 0 given by s(t) = t2 − 6t + 2 metres at time t seconds, t ≥ 0.Find the average velocity from t =1 to t = 3 seconds.

Question 6

A particle moves with velocity function v(t) = 8t − t2 cms−1, t ≥ 0.Find:(i) the velocity of the particle when t = 2 seconds(ii) the average acceleration from t =1 to t = 3 seconds.

8

SET 1 Derivative function and its graph

Question 1

Sketch the function f (x ) = 2. Given that m = f '(x ), find f '(x ).

Question 2

Sketch the function f (x ) = 2x + 3. Given that m = f '(x ), find f '(x ).

Question 3

For f (x ) = x 2 +1, find f '(x ) given that f '(x ) = lim

h→0

f (x +h)− f (x )h

.

9

SET 1

Derivative function and its graph

Question 4

For f (x ) = x 2 + x, find f '(3) given that f '(x ) = lim

h→0

f (x +h)− f (x )h

.

Question 5

For what value of x is the gradient of the tangent to the curve f (x ) =1− x 2 equal to -6?

Question 6

Sketch the graph of f (x ) = (x − 3)(x +1)in the domain − 2 ≤ x ≤ 4.From the graph find the coordinates ofthe stationary point.

10

SET 1

Derivative function and its graphQuestion 7

For the function f (x ) = x 2 find f '(x ).On separate graphs sketch f (x ) = x 2

and f '(x ) in the domain − 2 ≤ x ≤ 2.

Question 8

For the function f (x ) = −x 2 find f '(x ).On separate graphs sketch f (x ) = −x 2

and f '(x ) in the domain − 2 ≤ x ≤ 2.

Question 9

Consider the family of curves f (x ) = x 2 + c (c is a constant). Find f '(x )and discuss the relationship between f (x ) and f '(x ).

11

SET 1

Derivative function and its graphQuestion 10

For the function f (x ) = x 3 find f '(x ).On separate graphs sketch f (x ) = x 3

and f '(x ) in the domain − 2 ≤ x ≤ 2.

Question 11

(i) Show that 1

x +h− 1

x= −h

x(x +h).

(ii)For f (x ) = 1x

, find f '(x ) given that f '(x ) = limh→0

f (x +h)− f (x )h

.

Question 12

Consider the adjacent displacement-timegraph.(s(t) in cm and t in secs) From the graph find the instantaneousvelocity of the object for t =1 secondand t = 3 seconds.

12

SET 1

Calculating with derivatives

Question 1

Differentiate each of the following functions with respect to x.

(i) f (x ) = 3x 2 + 5x + 4 (ii) f (x ) = 25

x5 − 13

x 3 + x

Question 2

Differentiate each of the following functions with respect to x.

(i) f (x ) = 4x 2 + 3xx

(ii) f (x ) = 2x 2 − 8x + 2

Question 3

Differentiate each of the following functions with respect to x.

(i) f (x ) = x x (ii) f (x ) = x 4 +1

x 2

13

SET 1

Calculating with derivativesQuestion 4

Differentiate each of the following functions with respect to x.

(i) f (x ) = 3x 2 +1( )4 (ii) f (x ) = 1

2x 2 +1

Question 5

Use the chain rule dydx

= dydu

× dudx

to differentiate:

(i) y = 13x − 4

(ii) 9 − x 2

Question 6

Differentiate each of the following functions with respect to x.

(i) f (x ) = x +1( ) x + 3( )3 (ii) f (x ) = 3x 2 −1( )2 x + 3( )4

14

SET 1

Calculating with derivativesQuestion 7

Differentiate each of the following functions with respect to x :

(i) f (x ) = x +1x − 2

(ii) f (x ) = x

x 2 −1

Question 8

Find the equation of the tangent to the curve y = 2x 2 + 4x + 3at the point (1,9).

Question 9

Find the equation of the normal to the curve y = 1

x −1for x = 2.

15

SET 1

Calculating with derivativesQuestion 10

Determine the value of the constant k if the tangent to the curve f (x ) = kx − 2

,

at the point where x = 4, has a gradient of 12

.

Question 11

An object moves in a straight line with position given bys(t) = t3 −11t2 +12t m from O, where t is in seconds, t ≥ 0.(i) Find the velocity function v(t).(ii) Find the instaneous velocity at 2 seconds.

Question 12

An object moves in a straight line with velocity given byv(t) = 8t3 − 3t2 ms−1, where t is in seconds, t ≥ 0.(i) Find the acceleration function v '(t).(ii) Find the instaneous acceleration at 4 seconds.

16

SET 1 ANSWERSGradients of tangents 7) see graph for solution 1) continuous 8) see graph for solution

2) The function is discontinous at (2,4)

9) f '(x ) = 2x

3) The function is discontinous at x = 0

10) see graph for solution

4) 2 +h 11) (i) proof (ii) − 1

x 2

5) 2.5 12) t =1, v =10 cms−1,t = 3, v = −10 cms−1

6) 2 Calculating with derivatives

Difference quotients 1) (i) 6x + 5 (ii) 2x 4 − x 2 +1 1) m = 2, curve increasing at (1,3) 2) (i) 4 (ii) 2 2) A 3)

(i)

32

x (ii) 2x − 2x

3) Proof 4) (i) 24x(3x 2 +1)3 (ii) − 24x / (2x 2 +1)3

4) 2 5) (i) − 3/(3x − 4)2 (ii) − x / 9 − x 2 5) − 2 m/sec

6) (i) 2(x + 3)2(2x + 3) (ii) 4(3x 2 −1)(x + 3)3(6x 2 + 9x −1)

6) 4 cms−2

7) (i) − 3/(x − 2)2 (ii) −1/(x 2 −1) x 2 −1

Derivative function and its graph

8) y = 8x +1

1) f '(x ) = 0

9) y = x −1

2) f '(x ) = 2

10) k = −2

3) f '(x ) = 2x

11) (i) v(t) = 3t2 − 22t +12 (ii) − 20 ms−1

4) 7

12) (i) v '(t) = 24t2 − 6t (ii) 360 ms−2

5) x = 3

6) (1,−4)

17

SET 1 SOLUTIONS Gradients of tangents Question 1

Sketch the function f (x ) = x 2 − 4 in the domain − 3 ≤ x ≤ 3.State if this function is continous or discontinous, giving reasons.

The function is continous for all values of x.There are no points of discontinuity orasymptotes.

Question 2

Sketch the function f (x ) = x 2 − 4x − 2

in the domain − 3 ≤ x ≤ 3.

State if this function is continous or discontinous, giving reasons.

f (x ) = x 2 − 4x − 2

= (x − 2)(x + 2)(x − 2)

∴ f (x ) = x + 2, x ≠ 2The function is discontinous at (2,4)

Question 3

Sketch the function f (x ) = 12x

in the domain − 3 ≤ x ≤ 3.

State if this function is continous or discontinous, giving reasons.

The function is discontinous at x = 0.x = 0 is an asymptote.

18

SET 1 SOLUTIONS Gradients of tangents Question 4

The secant intersects the parabola y = x 2

at 1,1( ) and 1+h,(1+h)2( ).Find the gradient of the secant.

The gradient of the secant m

s= (1+h)2 −1

1+h −1= 1+ 2h +h2 −1

h= 2h +h2

h= 2 +h

Question 5

The secant intersect the parabola y = x 2

at 1,1( ) and 1.5,2.25( ).Find the gradient of the secant.

The gradient of the secant m

s= 2.25 −1

1.5 −1= 1.25

0.5= 2.5

Question 6

The tangent to y = x 2 touches the curve at (1,1) and intersects the

x axis at 12

,0⎛⎝⎜

⎞⎠⎟as shown in the

adjacent diagram.Find the gradient of the tangent.

The gradient of the tangent mt=

1− 0( )1− 1

2

⎛⎝⎜

⎞⎠⎟

=1÷ 1− 12

⎛⎝⎜

⎞⎠⎟=1÷ 1

2= 2

19

SET 1 SOLUTIONS Difference quotients Question 1

The tangent to the parabola y = 4x − x 2

touches the parabola at (1,3) and intersects the x axis at (-0.5,0).Find the gradient of the tangent andhence, or otherwise, determine if theparabola is increasing or decreasing.

mt= 3 − 0

1− (−0.5)= 3

1.5= 2

∴mt> 0 → the parabola is increasing.

Alternatively, from the graph, as x increases y increases.

Question 2

The tangent to the curve y = xtouches the parabola at (4,2) and intersects the y axis at (0,1).Find the gradient of the tangent anddetermine if the curve is:A) increasing at a decreasing rate.B) increasing at an increasing rate.C) decreasing at a decreasing rate.D) decreasing at an increasing rate.

mt= 2 −1

4 − 0= 1

4∴m

t> 0 → the parabola is increasing.

From the graph as x increases the gradient of the tangent decreases.∴ A

Question 3

P x, f (x )( ) and Q x +h, f (x +h)( ) are points on the the curve f (x ) = x 2 + x.

Show that the gradient of the secant PQ is 2x +1+h.

mpq

= f (x +h)− f (x )(x +h)− x

=(x +h)2 + x +h( ) − x 2 + x( )

h

= x 2 + 2xh +h2 + x +h − x 2 − xh

= 2xh +h2 +hh

= 2x +1+h

20

SET 1 SOLUTIONS Difference quotients Question 4

A(1,5) and B(3,9) are points on the curve x = 6t − t2, as shown in the adjacent diagram.Find the average velocity from A to B.

The average velocity is the gradient of the secant AB.

vAB

=x

2− x

1

t2− t

1

= 9 − 53 −1

= 2

Question 5

A particle moves in a straight line with a displacement from 0 given by s(t) = t2 − 6t + 2 metres at time t seconds, t ≥ 0.Find the average velocity from t =1 to t = 3 seconds.

vav

=s

2− s

1

t2− t

1

= s(3)− s(1)3 −1

s(3) = 32 − 6 × 3 + 2 = −7s(1) =12 − 6 × 1+ 2 = −3

vav

= −7 − (−3)3 −1

= −42

∴The average velocity is − 2 m/sec

Question 6

A particle moves with velocity function v(t) = 8t − t2 cms−1, t ≥ 0.Find:(i) the velocity of the particle when t = 2 seconds(ii) the average acceleration from t =1 to t = 3 seconds.

(i) v(2) = 8 × 2 − (2)2 =12 cms−1

(ii) aav

=v

2−v

1

t2− t

1

= v(3)−v(1)3 −1

v(3) = 8 × 3 − (3)2 =15v(1) = 8 × 1− (1)2 = 7

∴ aav

= 15 − 73 −1

= 82= 4

∴ The average acceleration is 4 cms−2

21

SET 1 SOLUTIONS Derivative function and its graph Question 1 Sketch the function f (x ) = 2. Given that m = f '(x ), find f '(x ).

From the graph it can be seenthat for f (x ) = 2, f '(x ) = 0.i.e. for all f (x ) = k, f '(x ) = 0.where k is a real constant.

Question 2 Sketch the function f (x ) = 2x + 3. Given that m = f '(x ), find f '(x ).

f (x ) = 2x + 3 is in the gradient formof a line.∴m = 2→ f '(x ) = 2i.e. for all f (x ) = mx +bf '(x ) = m, where m is a real constant.

Question 3

For f (x ) = x 2 +1, find f '(x ) given that f '(x ) = lim

h→0

f (x +h)− f (x )h

.

f '(x ) = limh→0

(x +h)2 +1− (x 2 +1)h

= limh→0

x 2 + 2xh +h2 +1− (x 2 +1)h

= limh→0

2xh +h2

h= lim

h→02x +h

∴ f '(x ) = 2x

22

SET 1 SOLUTIONS Derivative function and its graph Question 4

For f (x ) = x 2 + x, find f '(3) given that f '(x ) = lim

h→0

f (x +h)− f (x )h

.

f '(x ) = limh→0

(x +h)2 + x +h − (x 2 + x )h

= limh→0

x 2 + 2xh +h2 + x +h − (x 2 + x )h

= limh→0

2xh +h2 +hh

= limh→0

2x +h +1

∴ f '(x ) = 2x +1∴ f '(3) = 2 × 3 +1

= 7

Question 5

For what value of x is the gradient of the tangent to the curve f (x ) =1− x 2 equal to -6?

f '(x ) = limh→0

1− (x +h)2 − (1− x 2)h

= limh→0

1− x 2 − 2xh −h2 −1+ x 2

h

= limh→0

−2xh −h2

h= lim

h→0− 2x −h

∴ f '(x ) = −2x− 6 = −2x

∴ x = 3

Question 6

Sketch the graph of f (x ) = (x − 3)(x +1) in the domain − 2 ≤ x ≤ 4.From the graph find the coordinates of the stationary point.

The stationary point will occur when thegradient function is zero. This occurs at the vertex of the parabola.For y = (x − 3)(x +1)The x intercepts are x = 3 and x = −1.

∴ The axis of symmetry is x = 3 + −12

=1

When x =1, y = (1− 3)(1+1) = −4∴ The vertex is (1,−4)∴The stationary point is (1,−4).

23

SET 1 SOLUTIONS Derivative function and its graph Question 7

For the function f (x ) = x 2 find f '(x ).On separate graphs sketch f (x ) = x 2

and f '(x ) in the domain − 2 ≤ x ≤ 2.

f '(x ) = limh→0

(x +h)2 − x 2

h

= limh→0

x 2 + 2xh +h2 − x 2

h

= limh→0

2xh +h2

h= lim

h→02x +h

∴ f '(x ) = 2x

For x < 0, f (x ) is decreasing → f '(x ) < 0For x = 0, f (x ) = 0, f '(x ) = 0For x > 0, f (x ) is increasing → f '(x ) > 0

Question 8

For the function f (x ) = −x 2 find f '(x ).On separate graphs sketch f (x ) = −x 2

and f '(x ) in the domain − 2 ≤ x ≤ 2.

f '(x ) = limh→0

−(x +h)2 − (−x 2)h

= limh→0

−x 2 − 2xh −h2 + x 2

h

= limh→0

−2xh −h2

h= lim

h→0− 2x −h

∴ f '(x ) = −2x

For x < 0, f (x ) is increasing → f '(x ) > 0For x = 0, f (x ) = 0, f '(x ) = 0For x > 0, f (x ) is decreasing → f '(x ) < 0

Question 9

Consider the family of curves f (x ) = x 2 + c (c is a constant). Find f '(x )and discuss the relationship between f (x ) and f '(x ).

f '(x ) = limh→0

(x +h)2 + c − (x 2 + c )h

= limh→0

x 2 + 2xh +h2 + c − (x 2 + c )h

= limh→0

2xh +h2

h= lim

h→02x +h

∴ f '(x ) = 2x

For f (x ) = x 2 + c, f '(x ) = 2xAs c varies the gradient of f (x ) = x 2 + cwill remain the same for each valueof x for the family of f (x ) = x 2 + c.e.g. f (x ) = x 2 + 2→ f (2) = 6 → f '(2) = 4f (x ) = x 2 −1→ f (2) = 3 → f '(2) = 4f (x ) = x 2 +11→ f (2) =15 → f '(2) = 4etc

24

SET 1 SOLUTIONS

Derivative function and its graph Question 10

For the function f (x ) = x 3 find f '(x ).On separate graphs sketch f (x ) = x 3

and f '(x ) in the domain − 2 ≤ x ≤ 2.

f '(x ) = limh→0

(x +h)3 − x 3

h

= limh→0

x 3 + 3x 2h + 3xh2 +h3 − x 2

h

= limh→0

3x 2h + 3xh2 +h3

h= lim

h→03x 2 + 3xh +h2

∴ f '(x ) = 3x 2

For x < 0, f (x ) is increasing → f '(x ) > 0For x = 0, f (x ) = 0, f '(x ) = 0For x > 0, f (x ) is increasing → f '(x ) > 0

Question 11

(i) Show that 1

x +h− 1

x= −h

x(x +h).

(ii)For f (x ) = 1x

, find f '(x ) given that f '(x ) = limh→0

f (x +h)− f (x )h

.

(i) 1

x +h− 1

x= x

x(x +h)− (x +h)

x(x +h)

= x − (x +h)x(x +h)

= −hx(x +h)

(ii) f '(x ) = lim

h→0

1x +h

− 1x

h

= limh→0

−hx(x +h)

h..... from (i)( )

= limh→0

−hx(x +h)

÷h

= limh→0

−1x(x +h)

∴ f '(x ) = − 1

x 2

Question 12

Consider the adjacent displacement-timegraph. (s(t) in cm and t in secs) From the graph find the instantaneousvelocity of the object for t =1 secondand t = 3 seconds.

The gradient of the tangent at (1,15) is 10∴ the instantaneous velocity =10 cms−1

The gradient of the tangent at (3,15) is −10∴ the instantaneous velocity = −10 cms−1

25

SET 1 SOLUTIONS

Calculating with derivatives Question 1

Differentiate each of the following functions with respect to x.

(i) f (x ) = 3x 2 + 5x + 4 (ii) f (x ) = 25

x5 − 13

x 3 + x

(i) f (x ) = 3x 2 + 5x + 4f '(x ) = 2 × 3x + 5

= 6x + 5

(ii) f (x ) = 25

x5 − 13

x 3 + x

f '(x ) = 5 × 25

x 4 − 3 × 13

x 2 +1

= 2x 4 − x 2 +1

Question 2

Differentiate each of the following functions with respect to x.

(i) f (x ) = 4x 2 + 3xx

(ii) f (x ) = 2x 2 − 8x + 2

(i) f (x ) = 4x 2 + 3xx

= 4x 2

x+ 3x

x= 4x + 3

∴ f '(x ) = 4

(ii) f (x ) = 2x 2 − 8x + 2

= 2(x 2 − 4)x + 2

= 2(x − 2)(x + 2)x + 2

= 2x − 4∴ f '(x ) = 2

Question 3

Differentiate each of the following functions with respect to x.

(i) f (x ) = x x (ii) f (x ) = x 4 +1

x 2

(i) f (x ) = x x = x × x1 2

= x 3 2

∴ f '(x ) = 32

x1 2 = 32

x

(ii) f (x ) = x 4 +1

x 2= x 4

x 2+ 1

x 2

= x 2 + x −2

∴ f '(x ) = 2x − 2x −1 = 2x − 2x

26

SET 1 SOLUTIONS Calculating with derivatives Question 4

Differentiate each of the following functions with respect to x.

(i) f (x ) = 3x 2 +1( )4 (ii) f (x ) = 1

2x 2 +1

(i) f (x ) = 3x 2 +1( )4Using the composite function rule:

f (x ) = g(x )( )n → f '(x ) = n g(x )( )n−1× g '(x )

f '(x ) = 4 3x 2 +1( )3 × 6x

∴ f '(x ) = 24x 3x 2 +1( )3

(ii) f (x ) = 1

2x 2 +1= 2x 2 +1( )−

12

f '(x ) = − 12

2x 2 +1( )−32 × 4x

∴ f '(x ) = −2x 2x 2 +1( )−32

= −2x

2x 2 +1( )3

Question 5

Use the chain rule dydx

= dydu

× dudx

to differentiate:

(i) y = 13x − 4

(ii) 9 − x 2

(i) Let u = 3x − 4 → y = 1u

= u−1

dudx

= 3dydu

= −u−2 = − 1

u2

dydx

= dydu

× dudx

→ dydx

= − 1

u2× 3

∴dydx

= −3

(3x − 4)2

(ii) Let u = 9 − x 2 → y = u = u12

dudx

= −2xdydu

= 12

u−1

2 = 1

2 udydx

= dydu

× dudx

→ dydx

= 1

2 u× −2x

∴dydx

= −x

9 − x 2

Question 6

Differentiate each of the following functions with respect to x.

(i) f (x ) = x +1( ) x + 3( )3 (ii) f (x ) = 3x 2 −1( )2 x + 3( )4

(i) f (x ) = x +1( ) x + 3( )3Using the product rule:f '(x ) = u(x )×v '(x )+v(x )× u '(x )

Let u(x ) = x +1( ) and v(x ) = x + 3( )3∴u '(x ) =1 v '(x ) = 3(x + 3)2 × 1

= 3(x + 3)2

∴ f '(x ) = x +1( ) × 3(x + 3)2 + x + 3( )3 × 1= (x + 3)2 3 x +1( ) + (x + 3){ }= (x + 3)2 4x + 6{ }= 2(x + 3)2(2x + 3)

(ii) f (x ) = 3x 2 −1( )2 x + 3( )4Using the product rule:f '(x ) = u(x )×v '(x )+v(x )× u '(x )

Let u(x ) = 3x 2 −1( )2 and v(x ) = x + 3( )4∴u '(x ) = 2 3x 2 −1( ) × 6x v '(x ) = 4(x + 3)3 × 1

=12x 3x 2 −1( ) = 4(x + 3)3

f '(x ) = 3x 2 −1( )2 ×4(x + 3)3 + x + 3( )4 ×12x 3x 2 −1( )= 4 3x 2 −1( )(x + 3)3 3x 2 −1( ) + 3x(x + 3){ }= 4 3x 2 −1( )(x + 3)3(6x 2 + 9x −1)

27

SET 1 SOLUTIONS Calculating with derivatives Question 7

Differentiate each of the following functions with respect to x :

(i) f (x ) = x +1x − 2

(ii) f (x ) = x

x 2 −1

(i) Using the quotient rule:

For f (x ) = u(x )v(x )

f '(x ) = v(x )× u '(x )− u(x )×v '(x )

v(x )( )2Let u(x ) = x +1, v(x ) = x − 2∴ u '(x ) =1, v '(x ) =1

∴ f '(x ) = (x − 2)× 1− (x +1)× 1(x − 2)2

= −3

(x − 2)2

(ii) Let u(x ) = x, v(x ) = x 2 −1( )12

∴ u '(x ) =1, v '(x ) = 12

x 2 −1( )−12 × 2x

= x

x 2 −1

∴ f '(x ) =x 2 −1 × 1− x × x

x 2 −1

x 2 −1( )2

=x 2 −1( )2 − x 2

x 2 −1÷ x 2 −1( )2

= −1

(x 2 −1) x 2 −1

Question 8

Find the equation of the tangent to the curve y = 2x 2 + 4x + 3at the point (1,9).

dydx

= 4x + 4

At x =1dydx

= 8

y − y1= m(x − x

1)

y − 9 = 8(x −1)y − 9 = 8x − 8∴ y = 8x +1

Question 9

Find the equation of the normal to the curve y = 1

x −1for x = 2.

y = (x −1)−1

dydx

= −(x −1)−2 ×1 = − 1

(x −1)2

At x = 2, y =1 and dydx

= −1

∴ gradient of the normal =1y − y

1= m(x − x

1)

y −1 =1(x − 2)∴ y = x −1

28

SET 1 SOLUTIONS Calculating with derivatives Question 10

Determine the value of the constant k if the tangent to the curve f (x ) = kx − 2

,

at the point where x = 4, has a gradient of 12

.

f (x ) = k(x − 2)−1

f '(x ) = −k(x − 2)−2 ×1

= − k

(x − 2)2

At x = 4

− k

(4 − 2)2= 1

2

−k = 12× 4

∴k = −2

Question 11

An object moves in a straight line with position given bys(t) = t3 −11t2 +12t m from O, where t is in seconds, t ≥ 0.(i) Find the velocity function v(t).(ii) Find the instaneous velocity at 2 seconds.

(i) v(t) = 3t2 − 22t +12

(ii) v(2) = 3 × (2)2 − 22 × (2)+12= −20 ms−1

Question 12

An object moves in a straight line with velocity given byv(t) = 8t3 − 3t2 ms−1, where t is in seconds, t ≥ 0.(i) Find the acceleration function v '(t).(ii) Find the instaneous acceleration at 4 seconds.

(i) v '(t) = 24t2 − 6t

(ii) v(4) = 24 × (4)2 − 6 × (4)= 360 ms−2