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1/18

Topic : DIFFERENTIATION

Unit E : Determine the first derivative of composite function using chain rule.

Example Exercise

1. 2)2( += xy

)2(2

1)2(2 1

+=

+=

x

xdx

dy

a. 4)3( += xy

[4(x+3)3]

b. 5)2( += xy

[5(x+2)4]

c. 3)8( += xy

[3(x+8)2]

2. 2)23( += xy

)23(6

3)23(2 1

+=

+=

x

xdx

dy

a. 4)32( += xy

[8(2x+3)3]

b. 5)24( += xy

[20(4x+2)4]

c. 3)85( += xy

[15(5x+8)2]3. 2)2(3 += xy

)2(6

1)2(23 1

+=

+=

x

xdx

dy

a. 4)2(5 += xy

[20(x+2)3]

b. 5)24(3 += xy

[60(4x+2)4]

c. 3)82(2 += xy

[12(2x+8)3]

4.

2)2(

2

+

=

xy

3

3

3

2

)2(

4

)2(4

1)2(22

)2(2

+

=

+=

+=

+=

x

x

x

dx

dy

xy

a.

4)2(

5

+

=

xy

5

20

(x+2)[- ]

b.

5)2(

3

+

=

xy

6

15

(x+2)[- ]

c.

3)8(

2

+

=

xy

4

6

(x+8)[- ]

5.

2)2(5

2

+

=

xy

2)2(5

2

+= xy

1)2(252 3 += xy

3)2(5

4

+

=

xy

a.

3)5(4

3

+

=

xy

4

9

4(x+5)[- ]

b.

6)32(5

4

=

xy

7

24

5(2x-3)[- ]

c.

4)43(2

5

=

xy

5

10

(3x-4)[ ]

Differentiation 9

2/18

Topic : DIFFERENTIATION

Unit F : Determine the Gradient of a Tangent and a Normal at a point on a Curve.

Example 1 : 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=-2

Hence, the gradient of tangent at the point (-2,5)

5

7)2(6)2(6

2when766

2

2

2

=

+=

=+=

==

xxx

xwhendx

dymT

Example: Given )32()( = xxxf and the gradient

of tangent at point P on the curve

y = f(x) is 29, find the coordinates of thepoint P.

Solution: )(xfy =

y = 2x2 3x since f(x) = x(2x-3)

34 = xdx

dy

At point P, 29=dx

dy

4x 3 = 29

x = 8y = 104

The coordinates of P is (8 , 104)

(1) Given that the equation of a parabola is2241 xxy += , find the gradient of the tangent

to the curve at the point (-1,-3)

8=Tm

(2) Find the gradient of the tangent to the curve

( )( )32 += xxy at the point (3,6).

7=Tm

(3) Given that the gradient of the tangent at point P on

the curve ( )252 = xy is 4, find the coordinatesthe point P.

P(2 , 1)

(4) Given2

4)(

xxxf = and the gradient of tangent

is 28. Find the value of x.

32

=x

Differentiation 10

3/18

Topic : DIFFERENTIATION

Unit G : Determine the Equation of a Tangent and a normal at a Point on a Curve.

Example 1 :Find the equation of the tangent at the point (2,7)

on the curve 53 2 = xy

Solution:

126(2)dx

dy2,when x

6

532

===

=

=

xdx

dy

xy

=12

Equation of tangent is

( )11 xxmyy T = ( )

01712xy24127

2127

=+

=

=

xy

xy

Example 2 :Find the equation of the normal at the point x = 1

on the curve2324 xxy +=

Solution:

46(1)2dx

dy,1when x

62

3242

=+==

+=

+=

xdx

dy

xxy

1=

Nm

when x = 1 , y = 4 2(1) + 3(1)2= 5

Equation of normal is

( )11 xxmyy N =

( )

021-y4x

)1(1204

14

1

5

=+

=

=

xy

xy

(1) Find the equation of the tangent at the point (1,9)

on the curve ( )252 = xy

2112 += xy

(2) Find the equation of the tangent to the curve

( )( )112 += xxy at the point where itsx-coordinate is -1.

33 = xy

(3) Find the equation of the normal to the curve

232 2 += xxy at the point where its

x-coordinate is 2.

0225 =+ yx

(4) Find the gradient of the curve32

4

+

=

xy at the

point (-2,-4) and hence determine the equation of

the normal passing through that point.

0308;8 == yxmT

Differentiation 11

4/18

5/18

Topic : DIFFERENTIATION

Unit I : Determine the Types of Turning Points(Minimum and Maximum Points)

Example :Find the turning points of the curve

31812223

++= xxxy and determine

whether each of them is a maximum or aminimum point.

Solution:

31812223

=

++=

dx

dy

xxxy

At turning points, 0=dx

dy

( )( )3or x1

031034

018246

2

2

==

=

=+

=+

x

xxxx

xx

Substitute the values of x into

318122 23 ++= xxxy

When x = 1 , 113)1(18)1(12)1(2 23 =++=y

When x = 3 , 333)1(18)3(12)3(2 23 =++=y

Thus the coordinates of the turning points are

and

24122

2

= xdx

yd

When x = 1 , 01224)1(122

2

==

dx

yd

Thus , (3 , 33) is the point

(1) Find the coordinates of two turning points on the

curve ( )32 = xxy

(1 , 2) and (1 , 2)

(2) Determine the coordinates of the minimum point of

442 += xxy .

(2 , 0)

(3) Given 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 turningpoint is a maximum or minimum point.

min. point = (0 , 5) ; max. point = ),2(323

Differentiation 13

6/18

Topic : DIFFERENTIATION

Unit J : Problems of Rates of Change

(1) Given that xxy 23 2 = andxis increasing at a

constant rate of 2 unit per second, find the rate ofchange ofywhenx= 4 unit.

1

2

44

)2)(22(

22

2)4(6

4

26

23

2

=

=

=

=

=

=

=

=

=

sunit

dt

dx

dx

dy

dt

dy

dx

dy

xWhen

x

dx

dy

xxy

dt

dx

(2) Given that xxy =24 andxis increasing at a

constant rate of 4 unit per second, find the rate ofchange ofywhenx= 0.5 unit.

12 unit s1

(3) Given thatx

xv 19 = andxis increasing at a

constant rate of 3 unit per second, find the rate ofchange of vwhenx= 1 unit.

30 unit s1

(4) SPM 2004 (Paper 1 Question 21) [3 marks]Two variables,xandy, are related by the equation

.2

3x

xy += Given thatyincreases at a constant

rate of 4 unit per second, find the rate of change ofxwhenx=2.

58 unit s1

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

dt

dx

dx

dy

dt

dy= , where

dt

dy is the rate of change of

y anddt

dx is the rate of change of x.

Differentiation 14

7/18

(1) The area of a circle of radius rcm increases at aconstant rate of 10 cm

2per second. Find the rate of

change of rwhen r= 2 cm. ( Use = 3.142 )

1

2

7957.0

410

4

)2(2

2

2

10

=

=

=

=

=

=

=

=

=

scmdt

dr

dt

drdt

dr

dr

dA

dt

dA

dr

dA

cmrWhen

rdr

dA

rA

dt

dA

(2) The area of a circle of radius rcm increases at aconstant rate of 16 cm

2per second. Find the rate of

change of rwhen r= 3 cm. ( Use = 3.142 )

0.8487 cm s1

(3) The volume of a sphere of radius rcm increases at aconstant rate of 20 cm3per second. Find the rate ofchange of rwhen r= 1 cm. ( Use = 3.142 )

1.591 cm s1

(4) The volume of water , Vcm, in a container is given

by ,83

1 3hhV += where hcm 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]

65 cm s1

Differentiation 15

8/18

Example :

The above figure shows a cube of volume 729 cm.

If the water level in the cube, hcm, is increasing at

the rate of 0.8 cm s 1 , find the rate of increase of

the volume of water.

Solution :

Let each side of the cube bexcm.Volume of the cube = 729 cm

x = 729x = 9

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 of wateris 64.8 cms 1 .

(1) A spherical air bubble is formed at the base of apond. When the bubble moves to the surface of thewater, it expands. If the radius of the bubble is

expanding at the rate of 0.05 cm s1, find the rate

at which the volume of the bubble is increasingwhen its radius is 2 cm.

8.0 cm3s1

(2) If the radius of a circle is decreasing at the rate

of 0.2 cm s 1 , find the rate of decrease of thearea of the circle when its radius is 3 cm.

2.1 cm2s1

h cm

9cm

9cm

9cm

h cm

Chain rule

V = 9 x 9 x h = 81h

dh

dV=81

dt

dh =rate of increase of

the water level

= 0.8 cm s1

Differentiation 16

9/18

(3) The radius of a spherical balloon increases at the

rate of 0.5 cm s1. Find the rate of change in thevolume when the radius is 15 cm.

450 cm3

s

1

(4) The edge of a cube is decreasing at the rate of3 cm s1. Find the rate of change in the volumewhen the volume is 64 cm3.

144cm3

s

1

(5) Diagram 1shows 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 = 3.142; volume of cone hr2

3

1= )

6.631cm3s1

(6) Oil is poured into an inverted right circular cone ofbase radius 6 cm and height 18 cm at the rate of

2-13 scm . Find the rate of increase of the height of

water level when the water level is 6 cm high.( Use = 3.142 )

0.1591 cm s1

60 cm

Water

40 cm

Diagram 1

Differentiation 17

10/18

(7)

The above figure shows an inverted cone with aheight of 20 cm and a base-radius of 4 cm. Water

is poured into the cone at the rate of 5 cm s1but

at the same time, water is dripping out from the

cone due to a leakage a the rate of 1 cm s1.

(a) If the height and volume of the water at timets

arehcm and Vcm respectively, show that

.75

1 3hV =

(b) Find he rate of increase of the water level inthe cone at the moment the water level is 12 cm.( Use = 3.142 )

(b) 0.2210 cm s1

(8)

The above diagram shows a solid which consistsof a cuboid with a square base of side 6xcm,surmounted by a pyramid of height 4xcm. The

volume of the cuboid is 5832 cm.

(a) Show that the total surface area of the solid,

Acm, is given by .3888

96 2

xxA +=

(b) If the value ofx increasing at the rate

0.08 ,scm-1

find the rate of increase of the

total surface area of the solid at the instantx= 4.

42 cm2s1

4 cm

20 cm

hcm

B

H

F G

C

A

D

4xcm

6xcm

6xcm

V

E

Differentiation 18

11/18

Topic : DIFFERENTIATION

Unit K : Problems of Small Changes and Approximations

(1) Given that xxy 42 += , find the smallchange inywhenxincreases from 2 to 2.01.

08.0

)01.0)(8(

8

24)2(2

01.0

201.2

01.22

42

42

=

=+=

=

==

+=

+=

y

y

xdx

dyy

xwhendx

dy

x

x

xdx

dy

xxy

(2) Given that xxy 32 += , find the smallchange inywhenxincreases from 6 to 6.01.

0.15

(3) Given that xxy = 22 , find the smallchange inywhenxdecreases from 8 to 7.98.

0.62

(4) Given that xy 4= , finddx

dy.

Hence, find the small change inywhenxincreases

from 4 to 4.02.

02.0;2 == yxdx

dy

xdx

dyy

dx

dy

x

y

xxyy

inchangesmallinchangesmallwhere

==

Small Changes

xdx

dyy

yyy

original

original

+=

+=

new

Approximate Value

Differentiation 19

12/18

(1) Given the area of a rectangle , xxA 23 2 += ,wherex is the width, find the small change in the

area when the width decreases from 3 cm to 2.98cm.Answer :

12

2

scm4.0

)02.0)(20(

20

32)3(6

02.0

98.23

98.23

26

23

=

=+=

=

=

=

+=

+=

A

A

xdx

dAA

xwhendx

dA

x

x

xdx

dA

xxA

(2) A cuboid with square base has a total surface area,

xxA 43 2 = , wherex is the length of the side ofthe base. Find the small change in the total surfacearea when the length of the side of the basedecreases from 5 cm to 4.99 cm.

0.26 cm2

(3) The volume, Vcm3, of a cuboid with rectangular

base is given by xxxV 32 23 += , wherexcmis the width of the base. Find the small change in

the volume when the width increases from 4 cm to

4.05 cm.

1.75 cm3

(4) In a pendulum of lengthx meters, the period T

seconds is given as10

2 x

T = . Finddx

dT.

Hence, find the small change in Twhenxincreasesfrom 2.5 m to 2.6 m.

5010; == T

xdxdT

second

Differentiation 20

13/18

Example :

The height of a cylinder is three times its radius.Calculate the approximate increase in the total

surface area of the cylinder if its radius increases

from 7 cm to 7.05 cm.

Solution :

Let the total surface area of the cylinder be A cm.

A = Sum of areas of the top and bottom circularsurface + Area of the curved surface.

( )2

2

2

8

322

22

r

rrr

rhrA

=

+=

+=

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 total surface

area of the cylinder is 5.6 cm .

(1) 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.

7.2 cm2

(2) The volume of a sphere increases from

.cm290tocm288 33 Calculate the

721

cm

It is given that

h=3r

rdrdA

rA

16

8 2

=

=

New r (7.05)

Minus old r(7)

Substitute r with the

old value of r, i.e. 7

Differentiation 21

14/18

(1) 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

( )

.1165250

320

3

32

4

203.216

5

2

4

2

4

03.222

4,

2

4

03.2

4

5

5

555

+

+

==+=

xdx

dy

xandywherey

( )

0.13125

160

1

32

4

298.1165

24

1.984

(b)

55

+

+

+

+=

xdx

dyyy

yyy

originalnew

originalnew

(2) Given3

27

xy= , find the value of

dx

dywhenx= 3.

Hence, estimate the value of .03.3

273

1=dx

dy; 0.97

(3) Given4

32

xy= , find

dx

dy.

Hence, estimate the value of .99.1

324

5

128

xdx

dy = ; 2.04

Differentiation 22

15/18

(1) Given thatx

y

=2

20, find the approximate

change inxwhenyincreases from 40 to 40.5.

1601

(2)SPM 2003 (Paper 1 Question 16)

Given that ,52 xxy += use differentiation tofind the small change inywhenxincreases from 3to 3.01. [3 marks]

0.11

(3) Given3

3

4rv = , use the differentiation method

to find the small change in v when rincreases from

3 to 3.01.

27.0

(4) Given that ,53

xy= find the value of

dx

dywhen

x= 4.

Hence, estimate the value of

(a)( )302.45

(b)( )399.35

25615=

dx

dy; (a) 0.07930 ; (b) 0.07871

Differentiation 23

16/18

Topic : DIFFERENTIATION

Unit L : Problems of Maximum and Minimum Values of a Function.

Example 1 :

Given )540(2 xxL = , find the value ofxforwhichLis maximum.Hence, determine the maximum value ofL.

Solution:

27

10240

3

16540

3

16

max3

16

080

803

163080

3

16

01580,0

0)1580(

01580

0

3080

1580

540

)540(

2

max

2

2

2

2

2

2

2

2

32

2

=

=

=

dx

ydatx=

x= that satisfies 0=dx

dywill maximizeyif the value of 0

2

2

=

+=

=

=

=

==

=

+=

=

+=

+=

L

Limisewillx

dx

Ld

dx

Ld

x

xxKnowing

xx

xxxx

dx

dLwhen

xdx

Ld

xxdx

dL

xxL

xxL

(1) Given )128

(4 2

xxL += , find the value ofxfor

whichLis minimum.Hence, determine the minimum value ofL.

192,4 min == Lx

(2) Given )283(2

1 2 += xxy , find the value ofx

for whichy is minimum.Hence, determine the minimum value ofy.

35

min34 , == yx

(3) SPM 2003(Paper 1, No 15)

Given thaty=14x(5 x), calculate(a) the value ofxwhenyis a maximum(b) the maximum value ofy [3 marks]

(a) 2.5 ; (b) 87.5

Differentiation 25