2.16 ADD MATH MODULE 16 DIFFETENTIATION - … papers Add Maths/differentiation-… · ADDITIONAL MATHEMATICS MODULE 16 . 1 CHAPTER 9 : DIFFERENTIATION ... SPM Question 10 Assessment

DIFFERE

ADDITMATHE

MODU

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IONALMATICS

NTIATION

LE 16

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1

CHAPTER 9 : DIFFERENTIATION

ContentS

Concept Map 2

9.1 First Derivative for Polynomial Function 3

Test Yourself 1 3

9.2 First Derivative of a Productof Two Polynomials 4

Test Yourself 2 5

9.3 First Derivative of a Quotientof Two Polynomials 6

Test Yourself 3 7

9.4 First Derivative of a Composite Function 8

Test Yourself 4 9

SPM Question 10

Assessment 11 – 12

Answer 13

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DIFFERENTIATION TECHNIQUES

dx

dy


ky

dx

du

du

dy

dx

dy

dx

dy

2

naxy

v

uy

dx

dy

dx

dy


uvy

3

9.1 First Derivative for Polynomial Function

dx

dykky thenconstant,aiswhere,If.1

dx

dyaxy n then,If.2

TEST YOURSELF 1

1. y = 10

dx

dy=

2. y = 5x

dx

dy=

3. f (x) = -2 3xf ‘(x)=

4. y =x

7

dx

dy=

5.33

1)(

xxf

f ‘(x)=

6. xxy 24

dx

dy=

7.

x

xx

dx

d5

12

2

2

dx

dy

xxy )23(

Maths Tip

To differentiate means

to find )('or xfdx

dy

8.



Always changea fractional

function to thenegative indexbefore finding
differentiation

4

9. Given xxy 43 2 , find the value ofdx

dy

when x =2.

10. Given 21)( xxxf , find the value of

).1('and)0(' ff

9.2 First Derivative of a Product of Two Polynomials

Example :

Given that 322 22 xxy , finddx

dy.

Solution : left right

322 22 xxy

dx

dy= xxxx 43222 22

Keep Diff. Keep Diff.Left right right left

= xxxx 12444 33

= xx 88 3

dx

duv

dx

dvu

dx

dy

xvuuvy

then

,offunctionsbothareandwhere,If



5

TEST YOURSELF 2

1. 21 2 xxxy

dx

dy=

2. xxxdx

d2413 32 =

3. Given 4312 xxy . Finddx

dywhen

x= -1.

4. Given a curve xxxxf 221 ,

finda) f '(x),b) the value of f ' (2)



6

9.3 First Derivative of a Quotient of Two Polynomials

2then,If

vdx

dvu

dx

duv

dx

dy

v

uy

Example:

If13

52)(

2

x

xxf , find f’(x)

Solution:

2)(

bottom

bottom

diff

top

keep

top

diff

bottom

keep

xf

2

2

13

352413

x

xxx

= 2

22

13

156412

x

xxx

= 2

2

13

1546

x

xx

top

bottom



7

TEST YOURSELF 3

1.2

3 2

x

xy

dx

dy=

dx

dy

x

xxy

1

22

2

3. Given2

2

1

51)(

x

xxf

. Find the value of

f '(−2).

4. Given ,1

1

x

xxf find the value of x

when f '(x) =2

1 .

2.



8

9.4 First Derivative of a Composite Function

i) Using the chain rule

ii) Using the formula

Example :

.find,532Given that4

dx

dyxy

Solution :

Using the chain rule Using the formula

Let u = 3x – 5, then 42uy

Therefore, 38and3 udu

dy

dx

du

Hence,

3

3

3

5324

24

38

x

u

u

dx

du

du

dy

dx

dy

3

3

4

5342

3538

532

x

xdx

dy

xy

)('and)('where, xgdx

duuf

du

dy

dx

du

du

dy

dx

dy

)(,1

abaxndx

dybaxy

nn

Substituteu=3x - 5

2 x 4

The function inthe brackets is

kept



The function inthe brackets isdifferentiated.

Power is reduced by 1

9

6.

TEST YOURSELF 4

1. 42 3xxy

dx

dy=

2. 52 323 xy .

dx

dy=

3. y = 435

1

x

dx

dy=

4. Given 23py and p = 2x – 1.

Finddx

dyin terms of x.

5. Differentiate 74 23 xx with respect

to x.

.1when'ofvaluethe

find,3)(Given that222

xf

xxxf



10

SPM QUESTION

SPM 2004 (Paper 1 – Question 20)

Differentiate 4523 2 xx with respect to x.

[3 marks]

Solution:

SPM 2002 (Paper 2 )

Given 23

2

11 ttp . Find

dt

dpand hence, find the values of t when .1

dt

dp

[5 marks]Solution :



11

ASSESSMENT ( 30 minutes)

1. .dx

dyfind,2

15

xxy

2. .dx

dyfind,

4

32x

y

3. Find

2

2

x

xx

dx

d.

4. If 5334 xxy ,finddx

dy.



12

5. If23

12

x

xy ,find

dx

dy.

6. If 63 2)( xxf ,find )(' xf

7. If 8952 23 xxxy , find the value ofdx

dywhen 1x .



13

ANSWER

Test Yourself 1

4,1.10

8.9

26.8

52

4.7

18.6

1.5

7.4

63.

5.2

0.1

3

4

2

2

4

x

xx

x

x

x

x

x

Test Yourself 2

13)

166).4

23.

260x6.2

13.1

2

24

2

b

xxa

x

x

Test Yourself 3

1,3.4

9

163.

1

16.2

2

123.1

22

2

2

2

x

x

x-x

x

xx

Test Yourself 4

48.6

236112.5

1224.4

35

203.

3260.2

3324.1

63

5

42

32

xxx

x

x

xx

xxx

Assessment

77.

218.6

23

7.5

112.4

13.

2

3.2

15.1

532

2

2

3

2

xx

x

x

x

x

x

SPM Question

3

12,.2

52566.13

t

xxx



DIFFERE

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MODU

http://mathsmoz


IONALMATICS

14

NTIATION

LE 17

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15

CHAPTER 9

CONTENTS PAGE

9.1 Gradient of a tangent and a normal at a pointon a curve .

1-2

9.2 Equations of a tangent and normal to a curve. 3-4

9.3 Second derivative of y=f(x) 5-6

9.4 Turning point (minimum and maximumpoint)

7-8

SPM Questions 9

Assessment 10-11

Answers 12



CHAPTER 9 : Differentiation

Learning Outcomes: 2.7 Determine the gradient of a tangent at a point on acurve.

2.8 Determine the equation of tangent at a point on acurve.

2.9 Determine the equation of normal at a point on acurve.

3.1 Determine coordinates of turning point on a curve3.2 Determine whether a turning point is a maximum or

minimum point.

9.1 GRADIENT OF A TANGENT AND A NORMAL AT A POINT ON ACURVE

Gradient

dx

dymT

Tm

Nm1NT mm

Normal

tangent

y=f(x)


Gradient

16

T

Nm

m1


17

Example: Find the gradient of the tangent to the curve 5732 23 xxxy

at the point (-2,5)

Solution: 5732 23 xxxy

766 2 xxdx

dy

At point (-2,5), x=-2Hence, the gradient at the point (-2,5)

5

7)2(6)2(6

7662

2

xx

dx

dy

TEST YOURSELF(1)

1. Given that the equation of a

parabola is 2241 xxy

Find the gradient of the tangent tothe curve at the point (-1,-3)

2. Find the gradient of the tangent,

dx

dyto the curve 32 xxy

at the point (3,6)

3. Given that the gradient of thetangent at point P on the curve

252 xy is – 4, find the

coordinates the point P.

4. Given2

4)(

xxxf and the

gradient of tangent is 28. Find thevalue of x.



18

9.2 EQUATIONS OF A TANGENT AND NORMAL TO A CURVE

Example:Find the equation of the tangent atthe point (2,7) on the curve

53 2 xySolution:

126(2)dx

dy2,when x

6

53 2

xdx

dy

xy

Gradient of tangent, m T =12

Equation of tangent is

11 xxmyy T

01712xy

24127

2127

xy

xy

Example:Find the equation of the normal atthe point (2,7) on the curve

53 2 xySolution:

126(2)dx

dy2,when x

6

53 2

xdx

dy

xy

Gradient of normal,12

1Nm

Equation of normal is

11 xxmyyN

086-x12y

)2(18412

212

17

xy

xy

P(x,y)

tangent

normal

y=f(x)

x

y

11

:

xxmyy

Equation

dx

dym

T

T

11

:

1

xxmyy

Equation

mm

N

T

N



19

TEST YOURSELF(2)

1. Find the equation of the tangent atthe point (1,9) on the curve

252 xy

2. Find the equation of the tangentto the curve 112 xxy

at the point where its x-coordinateis -1.

3. Find the equation of the normal

to the curve 232 2 xxy at

the point where its x-coordinateis 2.

4. Find the gradient of the curve

32

4

xy at the point (-2,-4)

and hence determine the equationof the normal passing through thatpoint.



9.3 Second derivative of y=f(x)

f '(x) or

f ″(x) or

Example

Given that 5223)( xxf

Solution:

300

60(1)(5)(1)"

1-(6x)2x-60(3

18)2x-20(3

23(2x-320

34)23(20)("

2x-3-20x

4235)(

23)(

232

232

32

42

42

42

5

f

x

xxf

xxxf

xxf

http://mathsm


,fin

)

3

)

2

2

x

)(xfy

oz

dx

dy First derivative

20

d f”(x) and f”(1)

16

)20)(4(

2

32

x

xxx

2

2

dx

ydSecond derivative

ac.blogspot.com

21

TEST YOURSELF(3)

1. Given that 3642 23 xxxy ,

find2

2

dx

yd

2. Given that xxxf 3402)( 2 ,

find f ”(x)

3. Given that 5)14()( xxf ,

find f ”(0)4. Given that 22 13 ts ,calculate

the value of2

2

dt

sdwhen t=

2

1



22

9.4 Turning Point (Minimum and Maximum Point)

0dx

dy

Example :

Find the turning points of the curve 318122 23 xxxy and determine

whether each of them is a maximum or a minimum point.Solution:

31812223

dx

dy

xxxy

At turning points, 0dx

dy

3or x1

031

034

082462

2

x

xx

xx

xx

Substitute values of x into 318122 23 xxxy

When x=1 , 113)1(8)1(12)1(2 23 y

When x=3 , 33)1(8)3(12)3(2 23 y

Thus the coordinates of the turning points are

24122

2

xdx

yd

When x=1 , 01224)1(122

2

dx

yd,

Thus (1,11) is the point

When x=3, 0624)3(122

2

dx

yd

Thus , (3,3) is the point

Turningpoint

0dx

dy

Maximum point

Minimum point

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and

ogspot.com

23

TEST YOURSELF(4)

1. Find the coordinates of two turning points on the curve 32 xxy

2. Determine the coordinates of the minimum point of 442 xxy

3. 523

2 23 xxy is an equation of a curve. Find the coordinates of the

turning points of the curve and determine whether each of the turning pointis a maximum or minimum point.



24

SPM QUESTIONS:

1. SPM 2005 (Paper 1, No 19)

Given that 2

53

1)(

xxh , evaluate h ″(1) [ 4 marks]

2. SPM 2003(Paper 1, No 15)

Given that y=14x(5-x),calculatea) the value of x when y is a maximumb) the maximum value of y [3 marks]



25

ASSESSMENT:

1. If 432 xy , find

2

2

dx

yd

2. Find the gradient of the curve 113 xxy at the point where its

x-coordinates is2

1

3. Given that 722 xxy ,find the value of x if 812

22 y

dx

dyx

dx

ydx



26

4 Find the coordinates of the point on the curve 253 xy where the gradient

of its tangent is 6.

5. Find the gradient function of the curve23

5)(

xxf .Hence obtain the equation of

normal to the curve at the point which x-coordinate is -1.

6. A curve has a gradient function xpx 52 ,where p is a constant. The tangent to the

curve at point (1,4) is parallel to a straight line 013 xy .Find the

value of p



27

ANSWERS

TEST YOURSELF (1)1. 82. 73. P= (2,1)

4. x=3

2

TEST YOURSELF(2)

1. y+12x-21=02. y+3x+3=03. 5y+x-22=04. Gradient = -8

8y-x+30=0

TEST YOURSELF(3)1. 12x + 82. 160-36x3 -3204. -3

TEST YOURSELF(4)

1. (1,-2) and (-1,2)2. The minimum point is (2,0)

3. Turning points are (2,3

23 ) and

(0,-5)(0,-5) is a maximum point

(2,3

23 ) is a minimum point

SPM QUESTIONS:

1.8

27

2. a)2

5x , b)

2

175y

ASSESSMENT:

1. 23248 x

2. -1

3. 1,5

3x

4. ( 2,1)

5. 2

23

15

xy , 3y=5x +2

6. p = 2



DIFFERE

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MODU

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IONALMATICS

28

NTIATION

LE 18

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29

CHAPTER 9 : DIFFERENTIATION

Content page

Concept Map 2

9.1 Rates of Change 3

Test Yourself 1 4 – 5

9.2 Small Changes and Approximations 6 – 7

Test Yourself 2 8 – 9

SPM Question 10 – 11

Assessment 12 – 14

Answer 15



APPLICATION OF DIFFERENTIATION

Rate of change

dt

dx

dx

dv

dt

dv

Small changes

xdx

dyy

Maximminimu

Approximation


um and

30

m values


h cmh cm 9 cm

9 cm

h cm

9.1 Rates of Change

If y = f (x) and x = g(t), then using the chain ruledt

dx

dx

dy

dt

dy , where

dt

dyis the

rate of change of y anddt

dxis the rate of change of x.

Rate of change of the volume of water,

1364.8

0.881

scm

dt

dh

dh

dV

dt

dV

Hence, the rate of increase of the volume of1

Chain rule

dt

dh =

= 0.8 cm s1

V = 9

dh

dV =81

Sharpen Your Skills

9 cm


The above figure shows a cube of volume729 cm³. If the water level in the cube, h

cm, is increasing at the rate of 0.8 cm s 1 ,find the rate of increase of the volume ofwater.

Solution :

Let each side of the cube be x cm.

Volume of the cube = 729 cm³

x³ = 729

x = 9

31

water is 64.8 cm³ s .


rate of increace of

the water level

x 9 x h = 81h

32

TEST YOURSELF 1

1. A spherical air bubble is formed at the base of a pond. When the bubble moves to thesurface of the water, it expands. If the radius of the bubble is expanding at the rate of

0.05 cm s 1 , find the rate at which the volume of the bubble is increasing when itsradius is 2 cm.

Answer :

2. If the radius of a circle is decreasing at the rate of 0.2 cm s 1 , find the rate of decreaseof the area of the circle when its radius is 3 cm.

Answer :



3.

The above figure shows an in

4 cm. Water is poured into th

water is dripping out from th

(a) If the height and volum

respectively, show that

(b) Find he rate of increaselevel is 12 cm.

Answer :

h cm

http://ma


verte

e co

e con

e of

V

of th

4 cm

thsm

20 cm

33

d cone with a height of 20 cm and a base-radius of

ne at the rate of 5 cm³ s 1 but at the same time,

e due to a leakage a the rate of 1 cm³ s 1 .

the water at time t s are h cm and V cm³

.75

1 3h

e water level in the cone at the moment the water

ozac.blogspot.com

34

9.2 Small Changes and Approximations

Sharpen Your Skills

xdx

dyy

dx

dy

x

y

xx

yy

inchangesmall

inchangesmallwhere

Small Changes Approximate Value

2

2

2

8

322

22

r

rrr

rhrA

h=3r

http://mathsmozac.b


xdx

dyy

yyy

original

original

new

Approximate change in the total surface

area is A

2cm6.5

05.0716

705.716

r

rdr

dAA

dr

dA

r

A

Hence, the approximate increase in the totalsurface area of the cylinder is 5.6π cm² .

rdr

dA

rA

16

8 2

New r (7.05)Minus old r(7)

Substitute r with theold value of r, i.e. 7

The height of a cylinder is three times itsradius. Calculate the approximateincrease in the total surface area of thecylinder if its radius increases from 7 cmto 7.05 cm.

Solution :

Let the total surface area of the cylinderbe A cm².

A = Sum of areas of the top and bottomcircular surface + Area of the curvedsurface.

It is given that

logspot.com

Sharpen Your Skills

.1165250

320

3

32

4

203.216

5

2

4

03.2

455

160

1

32

4

298.116

5

2

4

1.98

4

(b)

55

xdx

dyyy

yyy

originalnew

originalnew

2

03.2

original

new

x

x

2

98.1

original

new

x

x


Given that5

4

xy , calculate the value of

dx

dyif x = 2. Hence estimate the values of

55 98.1

4(b)

03.2

4)a(

Solution :

16

520,2When

2020

44

6

6

6

5

5

xdx

dyx

xx

dx

dy

xx

y

xdx

dyy

yyya

original

original

)( new

35

0.13125


36

TEST YOURSELF 2

1. It is given that .2

20

xy

Find the approximate change in x when y increases from

40 to 40.5.

2. Given that ,5

3xy find the value of

dx

dywhen x = 4. Hence, estimate the value of

(a) 302.4

5(b)

399.3

5



37

3. A cube has side of 6 cm. If each of the side of the cube decreases by 0.1 cm, find theapproximate decrease in the total surface area of the cube.

4. The volume of a sphere increases from .cm290tocm288 33 Calculate theapproximate increase in its radius.



38

SPM QUESTION


Given that ,52 xxy use differentiation to find the small change in y when x increases

from 3 to 3.01.[3 marks]

Solution:


Two variables, x and y, are related by the equation .2

3x

xy

Given that y increase at a constant rate of 4 units per second, find the rate of change of xwhen x =2.

[3 marks]

Solution :



39


The volume of water , V cm³, in a container is given by ,83

1 3 hhV where h cm is the

height of the water in the container. Water is poured into the container at the rate of

.scm10 -13 Find the rate of change of the height of water, in ,scm -13 at the instant when its

height is 2 cm.[3 marks]

Solution :



40

ASSESSMENT

1. Given .3

4 3rv Use the differentiation method to find the small change in v

when r increases from 3 to 3.01.

Answer :

2. Given .27

3xy Find the value of

dx

dywhen x = 3.

Hence, estimate the value of .03.3

273

Answer :



41

60 cm

Water40 cm

Diagram 1

3. Diagram 1 shows a conical container with a diameter of 60 cm and height of 40 cm.

Water is poured into the container at a constant rate of 1 000 -13 scm .

Calculate the rate of change of the radius of the water level at the instant when the

radius of the water is 6 cm. (Use π = 3.142; volume of cone hr 2

3

1 )

Answer :



42

H

F G

C

A

4x cm

6x cm

6x cm

V

4.

E

The above diagram shows a solid whicside 6x cm, surmounted by a pyramid o5832 cm³.

(a) Show that the total surface area o

.3888

96 2

xxA

(b) If the value of x increasing at the

the total surface area of the solid

Answer :

D

http://mathsmoza


B

h consif heigh

f the so

rate 0.0

at the in

c.blogs

t
sts of a cuboid with a square base of4x cm. The volume of the cuboid is
lid, A cm², is given by

8 ,scm -1 find the rate of increase of

stant x = 4.

pot.com

43

ANSWER

Test Yourself 1

1

13

36

25.3

2.1.2

cm0.8.1

cmsdt

dh

dt

dA

s

Test Yourself 2

1. 0.005

2.256

15

dx

dy

(a) 0.07695(b) 0.07871

3. -7.2 cm²

4. cm72

1

SPM Question

1. 0.11

2. 1

5

8 unitsdt

dx

3. 18333.0 cmsdt

dh

Assessment

1. 0.36π

2. ,1dx

dy0.97

3. 6.631 cm s 1

4. 12cm42)( sb



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2.16 ADD MATH MODULE 16 DIFFETENTIATION - … papers Add Maths/differentiation-… · ADDITIONAL MATHEMATICS MODULE 16 . 1 CHAPTER 9 : DIFFERENTIATION ... SPM Question 10 Assessment