chapter 9- differentiation - · pdf fileadditional mathematics module form 4 chapter 9-...

Additional Mathematics Module Form 4

Chapter 9- Differentiation SMK Agama Arau, Perlis

Page | 105

CHAPTER 9- DIFFERENTIATION

9.1 LIMIT OF A FUNCTION

Example 1:

)2(lim2

+→

xx

= 2 + 2

= 4

Brief explaination:

y

4

2

0 2 x

1. If x is not 2 but 1.9, 1.99 and 1.999, the value of y gets nearer and nearer to 4 but it does not exceed

4.

2. It is also just the same if the value of x is 2.1, 2.01, 2.001, the value of y gets nearer and nearer to 4

but it does not exceed 4.

3. In this situation, we can say that 2+x approaches 4 as x approaches 2 or 4)2(lim2

=+→

xx

4. Hence, the function 2+= xy has limit 4 as 2→x .

Example 2:

2

4lim

2

2 −

−

→ x

x

x

)2(

)2)(2(lim

2 −

−+

→ x

xx

x

)2(lim2

+→

xx

22 +=

4=

2+= xy

First of all, factorize the numerator first.

Then simplify.



Page | 106

Example 3:

)1

(limx

x

x

+

∞→

)1

(limxx

x

x+=

∞→

)1

1(limxx

+=∞→

01+=

1=

EXERCISE 9.1

Find:

(a) 2

223

limx

x

x

+

∞→

(b) )32(lim1

+→

xx

(c) 2

52lim x

x→

(d) 3

9lim

2

4 −

−

→ x

x

x

9.2 FIRST DERIVATIVE OF A FUNCTION

9.21 GRADIENT OF THE TANGENT TO A CURVE AND THE FIRST DERIVATIVE OF A FUNCTION

1. Tangent to a curve is a line which is just touching the curve and not cut the curve.

2. The gradient of the tangent to a curve can be determined by finding the small changes in y divided by

small changes in x.

Brief explaination:

Q(1.1,2.4)

P(1,2)

1. When the point Q is move nearer and neared to the point P, there will be a point which is very near to

point P but not the point P and there is a very small change in value of x and y at the point from point P.

2. The point and the point P are joined in a line that is the tangent of the curve.

We know that any number that divided by zero will

result infinity. For example:

∞=0

1. So if we can change the equation to be like this

01

=∞

that is 1 divided by will result zero.

For this question, at first we have to separate the terms

into two fractions. Then simplify for the terms that can

be simplified.

Substitute x= ∞ into x

1and becomes

∞

1. We know that

01

=∞

.

o

o



Page | 107

2. The small change in y is the difference in value of y between the point and point P while the small

change in x is the difference between the value of x of the point and point P.

3. Small changes in y can be written as yδ that is read as “delta y” while small changes in x can be

written as xδ that is read as “delta x”.

Q(x+ xδ , y + yδ )

yδ

P(x, y) xδ x+ xδ

4. From the graph above, we know that x

ymPQ

δ

δ= .

5. But at point P, x

ym

xP

δ

δδ 0lim

→=

6. x

y

x δ

δδ 0lim

→is the gradient of tangent at point P.

7. To write x

y

x δ

δδ 0lim

→, it is a quite long, so we can write them as

dx

dy.

8. dx

dyis differentiation.

9.22 DIFFERENTIATION BY THE FIRST PRINCIPLE

Q( x+ xδ , y + yδ )

P(x, y)

2xy =

22

2

2

)(

xxxxyy

xxyy

δδδ

δδ

++=+

+=+

2xy =

1

2

The both points lie on the same curve, so

they can be solved simultaneously.

Tangent to the curve

This graph show the point Q that

very close to point P that has moved

nearer to point P

o

o

o

o



Page | 108

- ,

xxx

y

xxx

y

x

xxx

x

y

xxxy

xxxxxyyy

xxδ

δ

δ

δδ

δ

δ

δδ

δ

δ

δδδ

δδδ

δδ+=

+=

+=

+=

−++=−+

→→2limlim

2

2

2

2

00

2

2

222

xdx

dy2=

EXERCISE 9.20

Find dx

dyof the following equation using the first principle.

(a) 2

3xy = (b) 522 += xy

(c) x

y2

= (d) x

y3

5 −=

(e) 2

31 xy −= (f) xxy 532 −=

- Finding dx

dyby using formula -

naxy =

Given that y = f(x),

The first derivate dx

dyis equivalent to )(

' xf ,

That is ifnaxxf =)( ,

Then,

)()( ' xf

dx

xdf=

2 1

1..

−= nxnadx

dy

Left hand side and right hand side are

divided by xδ to make the termx

y

δ

δ. The

right hand side is then simplified.

For left hand side, we already know that

x

y

x δ

δδ 0lim

→ can be converted into

dx

dy. For

right hand side, the value of xδ is

replacing by zero.

When Q approaches P, there is no

change in x or xδ approaches 0.

1..)('

−= nxnaxf

If the 2xy = , then x

dx

dy2= .

If2

)( xxf = , then xxf 2)(' =

If xy 2=

2

)2()(

=

=

dx

dy

xdx

dy

dx

d

If xxf 2)( =

2)('

2)(

)2()]([

=

=

=

xf

dx

xdf

xdx

dxf

dx

d



Page | 109

Example 1:

2xy =

)(2x

dx

d

dx

dy=

12.2.1

−= xdx

dy

xdx

dy2=

Example 2:

xxf

xxxf

xxf

6)('

.0.6.2.3)('

63)(

1012

2

=

+=

+=

−−

EXERCISE 9.21

1. Find dx

dyfor each of the following equation. Hence, find the value of

dx

dyat point where 2=x .

(a) 424 −= xy

(b) )5(2 += xxy

(c) 2

22

x

xxy

+=

2. Find )(' xf for each of the following functions. Hence, find the value of )3('f .

(a) xxxf 3)(2 +=

(b) 2

)12()( −= xxf

(c) )3)(22()( −−= xxxf

9.3.1 Differentiate expression with respect to x

Example 1:

Differentiate 42

1 4 −x with respect to x.

Solution:

)42

1(

04 xxdx

d−

10144.0

2

1.4

−− −= xx

32x=

We add the terms x0 because to do

differentiation, it involves x if it is respect

to x. x0 = 1.

When differentiate the terms without

the variable or unknown, it will result

zero.

Tips…

Compare the solution by using formula

and by using the first principle. We get

the same answer for the same equation.

We cannot change the expression into an

equation and then differentiate it.

The solution must be started with

)42

1(

4 −xdx

d, cannot with 4

2

1 4 −x .



Page | 110

Example 2:

Differentiate )32)(32( +− xx with respect to x.

x

xx

xxdx

d

xxdx

d

8

.9.0.4.2

)94(

)]32)(32[(

1012

02

=

−=

−

+−

−−

EXERCISE 9.22

Differentiate each of the following with respect to x.

(a) 532 −x

(b) x

x22 −

(c) )43)(43( +− xx

9.3 FIRST DERIVATIVE OF COMPOSITE FUNCTION

Example 1:

Find dx

dyfor the function

2)12( += xy

Solution:

Method 1

2)12( += xy

1442 ++= xxy

Method 2

2)12( += xy

Let

12 += xu

2=dx

du

2uy =

First of all, expand the bracket. Then

differentiate the expression.

We add the terms x0 because to do

differentiation, it involves x if it is respect

to x. x0 = 1.

48 += xdx

dy

We know the value du

dyand

dx

du, but we

have to find dx

dy. So we have to use the

concept below.

First of all, expand the bracket. Then

differentiate it



Page | 111

udu

dy2=

22 ×= udx

dy

u4=

)12(4 += x

48 += x

Brief Explaination:

From the second method above, the concept to differentiate it is:

1. Consider the expression in the bracket as a term such as u but cannot x. Then differentiate it.

2uy =

udu

dy2=

2. Then, differentiate the part in the bracket.

12 += xu

2=dx

du

3. Multiply both of them and it will resultdx

dy.

4. We can use the method 2 in all situations easily but method 1 can be used easily in certain situations

only that is the number of power is small such as 2.

Example 2:

Find dx

dyfor the function

5)43( += xy

Let

43 += xu

3=dx

du

5uy =

45u

du

dy=

dx

du

du

dy

dx

dy×=

If we simplify dx

du

du

dy

dx

dy×= at right

hand side, it will result dx

dy. This is called

chain rule.

We can use method 1 to differentiate if the

number of power is small such as 2. But, if it is

the power of 4, 5 and above, we can still use

the method 1 but it is very complicated to

solve but it is easy to differentiate by using the

method 2.

Tips…

We are not supposed to use method 1. It is

quite complicated have to expand the bracket

because it is the power of 5. So we can use the

method 2.



Page | 112

354 ×= u

dx

dy

4

15u=

4

)43(15 += x

We can also write directly like this:

3.)43(515−+= x

dx

dy

4

)43(15 += x

EXERCISE 9.3

Find dx

dyfor each of the following equation.

(a) 4

)31( xy −=

(b) 3

)1

1(x

y +=

(c) 4

)53(

1

−=

xy

9.4 FIRST DERIVATIVE OF TWO POLYNOMIALS

9.41 FIRST DERIVATIVE OF THE PRODUCT OF TWO POLYNOMIALS

)1)(12( ++= xxy

u v

uvy =

vuuvvuuvyy

vvuuyy

δδδδδ

δδδ

+++=+

++=+ ))((

Substitute 43 += xu into 4

15u

dx

du

du

dy

dx

dy×=

2

1 How to get the formula?



Page | 113

- , ,

)(limlim00 x

vu

x

uv

x

vu

x

y

x

vu

x

uv

x

vu

x

y

x

vuuvvu

x

y

vuuvvuy

uvvuuvvuuvyyy

dxdx δ

δδ

δ

δ

δ

δ

δ

δ

δ

δδ

δ

δ

δ

δ

δ

δ

δ

δδδδ

δ

δ

δδδδδ

δδδδδ

++=

++=

++=

++=

−+++=−+

→→

x

vu

x

uv

x

vu

x

y

dxdxdxdxdxdxdx δ

δδ

δ

δ

δ

δ

δ

δ0000000

limlimlimlimlimlimlim→→→→→→→

×××××=

dx

dv

dx

duv

dx

dvu

dx

dy.0++=

Example:

Find dx

dyfor the function )1)(12( ++= xxy

)1)(12( ++= xxy

u v

1

1

2

12

=

+=

=

+=

dx

dv

xv

dx

du

xu

The formula is dx

duv

dx

dvu

dx

dy+= , just substitute each term into the formula to find

dx

dy

)2)(1()1)(12( +++= xxdx

dy

34

2212

+=

+++=

x

xx

2 1

dx

duv

dx

dvu

dx

dy+=



Page | 114

EXERCISE 9.40

1. Find dx


dx

dyat point where 1−=x .

(a) )2)(31( +−= xxy

(b) 2

)3)(2( −+= xxy

(c) 32

)2()3( ++= xxy

2. Find )(' xf for each of the following functions. Hence, find the values of )1('f and 2 )2(' −f

(a) )3)(52()( ++= xxxf

(b) 5

)4)(12()( +−= xxxf

(c) 24

)3()22()( −−= xxxf


Example:

Differentiate 42

)52(3 −xx with respect to x.

42)52(3 −xx

u v

2.)52(4

)52(

6

3

3

4

2

−=

−=

=

=

xdx

dv

xv

xdx

du

xu

3

)52(8 −= x

)56()52(6

)]52(4[)52(6

)52(6)52(24

6.)52()52(8.3

])52(3[

3

3

432

432

42

−−=

−+−=

−+−=

−+−=

−

xxx

xxxx

xxxx

xxxx

xxdx

d

If it is given an equation, the formula used is

dx

duv

dx

dvu

dx

dy+= . If it is given an expression not

equation, the formula used is just

dx

duv

dx

dvu + without equal sign because it is an

expression not an equation.

Factorize



Page | 115

EXERCISE 9.41


(a) 3

)42)(53( xx −−

(b) 53

)65(4 −xx

(c) 43

)4()98( +− xx

9.42 FIRST DERIVATIVE OF THE QUOTIENT OF TWO POLYNOMIALS

1

12

+

+=

x

xy

v

uy =

vv

uuyy

δ

δδ

+

+=+

- , ,

+

−

=

+

−

=

×+

−=

+

−=

+

−−+=

+

+−

+

+=

−+

+=−+

→→ )(limlim

)(

1

)(

)(

)(

)(

)(

)(

)(

00 vvv

x

vu

x

uv

x

y

vvv

x

vu

x

uv

x

y

xvvv

vuuv

x

y

vvv

vuuvy

vvv

vuuvuvuvy

vvv

vvu

vvv

uuvy

v

u

vv

uuyyy

dxdx δδ

δ

δ

δ

δ

δ

δδ

δ

δ

δ

δ

δ

δδ

δδ

δ

δ

δ

δδδ

δ

δδδ

δ

δ

δ

δδ

δ

δδ

u

v

2

2 1

1 How to get the formula?



Page | 116

)(limlim

limlimlimlim

)(lim

00

0000

0

vvv

x

vu

x

uv

vvv

x

vu

x

uv

dxdx

dxdxdxdx

dx

δδ

δ

δ

δ

δδ

δ

δ

δ

+×

×−×

=

+

−

=

→→

→→→→

→

vv

dx

dvu

dx

duv

dx

dy

.

−

=

Example:

Find dx

dyfor the function

1

12

+

+=

x

xy by using formula

2v

dx

dvu

dx

duv

dx

dy−

= .

1

1

2

12

=

+=

=

+=

dx

dv

xv

dx

du

xu

Substitute each term into the formula to finddx

dy,

2)1(

)1)(12()2)(1(

+

+−+=

x

xx

dx

dy

2

2

2

)1(

3

)1(

1222

)1(

)12()1(2

+=

+

+−+=

+

+−+=

x

x

xx

x

xx

2v

dx

dvu

dx

duv

dx

dy−

=



Page | 117

EXERCISE 9.42

1. Find dx


dx

dyat point where 1=x .

(a) ( )

52

232

−

+=

x

xy

(b) ( )( )2

3

1

12

+

+=

x

xy

(c) ( )( )3

2

74

65

−

−=

x

xy

2. Find )(' xf for each of the following functions. Hence, find the values of )1('f and 2 )5('f

(a) 52

2)(

2

−

+=

x

xxxf

(b) ( )

52

233

−

+=

x

xy

(c) ( )

xx

xy

52

232

2

−

+=


Example:

Differentiate 1

232

+

+

x

xxwith respect to x.

1

1

26

232

=

+=

+=

+=

dx

dv

xv

xdx

du

xxu

+

+

1

232

x

xx

dx

d

2

2

)1(

)1)(23()26)(1(

+

+−++=

x

xxxx

2

22

)1(

23286

+

+−++=

x

xxxx

If it is given an equation, the formula used is

2v

dx

dvu

dx

duv

dx

dy−

= . If it is given an expression not

equation, the formula used is just

2v

dx

dvu

dx

duv −

without equal sign because it is an

expression not an equation.



Page | 118

2

2

)1(

2103

+

++=

x

xx

EXERCISE 9.43


(a) 23

32

−

+

x

x

(b) 2

34

x

xx +

(c) 12

52

+x

x

9.5 TANGENT AND NORMAL TO THE CURVE

1. Tangent to a curve is a line which is just touching the curve and not cut the curve.

2. Normal to a curve is a straight line perpendicular to the tangent at same point on the curve as shown

in the figure above. Hence, since the both straight lines are perpendicular to each other,

1tan −=× normalgent mm

3. We have learned that dx

dyis the gradient of tangent. So, when we are going to find the gradient of

normal, at first we have to find the gradient of tangent.

4. If it is given the gradient of normal, we can find the gradient of the tangent by using formula above.

Normal to the curve

Tangent to the curve



Page | 119

Example 1:

Find the equation of the tangent and the equation of normal to the curve 5922 −−= xxy at the point

(3, 4)

Solution:

94

5922

−=

−−=

xdx

dy

xxy

At point (1, 3),

9)3(4 −=dx

dy

3

912

=

−=

Gradient of tangent is 3.

3

1

13

1tan

−=

−=×

−=×

normal

normal

normalgent

m

m

mm

The equation of the tangent at (3, 4) is

63

933

33

3

−=

−=−

=−

−

xy

xy

x

y

The equation of the normal at (3, 4) is

0123

393

3

1

3

3

=−+

−=−

−=−

−

yx

xy

x

y

We have learned that to find the gradient, we

use the formula

12

12

2−

−

x

yy. Use the point (x, y)

that is as a general point and given point (3, 4).

Substitute the value of x into the equation to find the

gradient of tangent at point (1, 3)



Page | 120

Example 2:

Find the equation of normal to the curve 2432 +−= xxy that is parallel to the line 52 =− xy .

Solution:

46

2432

−=

+−=

xdx

dy

xxy

2

5

2

1

52

+=

=−

xy

xy

m = 2

1

normalm is parallel to this line, so

2

1=normalm .

2

12

1

1

tan

tan

tan

−=

−=×

−=×

gent

gent

normalgent

m

m

mm

We know that 46 −= xdx

dyand 2tan −=gentm , hence

3

1

246

=

−=−

x

x

2tan −=gentm and 2

1=normalm at point which

3

1=x .

2432 +−= xxy

3

1=x ,

23

14

3

13

2

+

−

=y

1=

We do not know any point so we cannot find

the gradient of tangent at the moment .

Given that the gradient of normal is equal to

the gradient of this line. So we can find the

gradient of normal.

Normal to a curve is a straight line

perpendicular to the tangent. We know the

gradient of normal, so we can find the gradient

of tangent.

At the early of the solution, we got

46 −= xdx

dythat is the gradient of tangent. From

the gradient of normal, we got the gradient of

tangent. So compare these two equations

We are going to find the value of y for this point

that results the gradient of tangent and normal are

-2 and 2

1respectively. We substitute the value of

x into the equation of the curve that is

2432 +−= xxy to find the value of y.



Page | 121

Hence the point is

1,

3

1.

The equation of the normal at

1,

3

1 is

0563

1366

3

122

2

1

3

1

1

=+−

−=−

−=−

=

−

−

yx

xy

xy

x

y

EXERCISE 9.5

Find the equation of tangent and the equations of normal for each of the following functions at given

points:

(a ) xxxy 4623 +−= ; 3=x

(b)2

)3)(2( −+= xxy ; 2−=x

(c)23

32

−

+=

x

xy ; 1=x

9.6 SECOND ORDER DIFFERENTIATION

1. First differentiation is dx

dy.

2. Second order differentiation is 2

2

dx

ydthat is we differentiate for the second times.

Example 1:

Given xxxy 4623 +−= ,find

2

2

dx

yd.

126

4123

46

2

2

2

23

−=

+−=

+−=

xdx

yd

xxdx

dy

xxxy

Multiply the equation by 3

Differentiate one more time

dx

d

dx

dy=

2

2

dx

yd



Page | 122

Example 1:

Givenx

xxxf1

4)(3 ++= ,find )(

'' xf .

3

''

3''

22'

13

3

26)(

26)(

143)(

4)(

14)(

xxxf

xxxf

xxxf

xxxxf

xxxxf

+=

+=

−+=

++=

++=

−

−

−

EXERCISE 9.6

1. Find 2

2

dx

ydfor each of the following equation. Hence, find the value of

2

2

dx

ydat point where 1=x .

(a) xxy 524 −=

(b) )5(22 xxxy +=

(c) 2

232

x

xxy

+=

2. Find )('' xf for each of the following functions.

(a) xxxf 3)(2 +=

(b) 2

)23()( −= xxf

(c) 3

)3)(24()( −−= xxxf

9.7 CONCEPT OF MAXIMUM AND MINIMUM VALUES

y

C

D

A

B

x

1. Based on the graph above:

(a) A and C are maximum points

(b) B and D are minimum points

(c) A, B, C and D are turning points

If )()( ' xf

dx

xdf= ,

then )()( ''

2

2

xfdx

xfd=




Page | 123

2. At turning points, 0=dx

dy

3. This is because the line of the tangent at the turning points is a horizontal line that is parallel to x-axis

which is the gradient is 0. dx

dyis the gradient of tangents so at at turning points, 0=

dx

dy.

9.6.1 Steps to determine a turning point is maximum or minimum

1- Find dx

dy

2- Determine the turning points, find the value of x then find the value of y

3- Determine whether the turning point (x, y) is maximum or minimum.

(a) minimum point:

x <x1 x1 >x1

dx

dy negative 0 positive

Sketch position of

tangent

Shape of the graph

(b) maximum point:

x <x1 x1 >x1

dx

dy positive 0 negative

Sketch position of

tangent

Shape of the graph

Example 1:

Find the turning point for the function 59323 +−−= xxxy . Hence, state each point is minimum or

maximum.

Solution:

Method 1

963

593

2

23

−−=

+−−=

xxdx

dy

xxxy



Page | 124

At turning point, 0=dx

dy.

0)3)(1(

032

0963

2

2

=−+

=−−

=−−

xx

xx

xx

1−=x or 3=x

(i) 1−=x

5)1(9)1(3)1(23 +−−−−−=y (-1, 10)

10=

(ii) 3=x

5)3(9)3(3)3(23 +−−=y (3, -22)

2=y

The turning points are (-1 , 10) and (3,-22)

(i)

(i) 2−=x

15

9)2(6)2(32

=

−−−−=

dx

dy

dx

dy

(i) 0=x

So, the point (-1, 10) is a maximum point.

(ii)

(i) 2=x

9

9)2(6)2(32

=

−−=

dx

dy

dx

dy

(i) 4=x

So, the point (3, -22) is a minimum point.

x <-2 -1 0

dx

dy positive 0 negative

Sketch position of

tangent

Shape of the graph

x 2 3 4

dx

dy negative 0 positive

Sketch position of

tangent

Shape of the graph

9

9)0(6)0(32

−=

−−=

dx

dy

dx

dy

15

9)4(6)4(32

=

−−=

dx

dy

dx

dy

Factorize to find the values of x

Substitute the value of x into the equation of the curve

that is 59323 +−−= xxxy to find the value of y.

Use the first value of x that is -1. Take a value that

is greater than -1 such as 0 and a value that is less

than -1 such as -2. Then solve it to find out either it

is a maximum or minimum point. For the second

value of x that is 3, repeat the same steps to find

out either it is a maximum or minimum point.



Page | 125

Method 2

963

593

2

23

−−=

+−−=

xxdx

dy

xxxy

At turning point, 0=dx

dy.

0)3)(1(

032

0963

2

2

=−+

=−−

=−−

xx

xx

xx

1−=x or 3=x

(i) 1−=x

5)1(9)1(3)1(23 +−−−−−=y (-1, 10)

10=

(ii) 3=x

5)3(9)3(3)3(23 +−−=y (3, -22)

2=y

The turning points are (-1 , 10) and (3,-22).

9632 −−= xx

dx

dy

662

2

−= xdx

yd

At point (-1, 10),

6)1(62

2

−−=dx

yd

12−=

02

2

<dx

yd

(-1, 10) is a maximum point

At point (3, -22),

6)3(62

2

−=dx

yd

12=

To find the coordinates of the turning

points. Substitute the value of x into the

equation 59323 +−−= xxxy to find

the value of y.

Factorize to find the values of x


If 02

2

<dx

yd, then it is a maximum value. If 0

2

2

>dx

yd,

then it is a minimum value.

Tips…



Page | 126

02

2

>dx

yd

(3, -22) is a minimum point.

EXERCISE 9.7

1. Find the turning point for each of the following function. Hence, state each point is minimum or

maximum.

(a )x

xy1

+=

(b)2

2 16)(

xxxf +=

2. Find the turning point at the following curve

(a )2

8 xxy −=

(b ) 2

2 962)(

xxxf +=

9.8 RATE OF CHANGE

1. The first derivative for a function denotes the change in the quantity y with respect to the change in

the quantity x.

2. Rate of change is differentiation that is respect to time

dt

d

4. The formula for rate of change is

where A and B are variables that can be changes depends on the case and situation.

Example 1:

The sides of a cube increases at the rate of 1.4cms-1

. Find the rate of change of the volume when the

sides measure 5 cm.

Solution:

In this case, A is the volume (V) and B is the side(s).

Given that 1

4.1−= cms

dt

ds, cms 5= and

3sV = . We are going to find

dt

dV.

2

3

3sds

dV

sV

=

=

dt

dB

dB

dA

dt

dA×=

3sV = is the formula for the volume of

cube. Differentiate it that is respect to s.



Page | 127

The formula is

dt

ds

ds

dV

dt

dV×=

Substitute2

3sds

dV= into the formula.

dt

dss

dt

dV×= )3(

2

When 1

4.1−= cms

dt

ds, 5=s ,

4.1)5(32 ×=

dt

dV

13

105

4.175

−=

×=

scm

Example 2:

The volume of a sphere decreases at the rate of 13

4−

scm . Find the rate of change of the radius of

sphere when the radius is 3cm.

Solution:

In this case, A is the volume (V) and B is the radius (r).

Given that 13

4−−= scm

dt

dV, cmr 3= and

3

3

4rV π= . We are going to find

dt

dr.

2

3

4

3

4

rdr

dV

rV

π

π

=

=

The formula is

dt

dr

dr

dV

dt

dV×=

Substitute 2

4 rdr

dVπ= into the formula,

dt

drr

dt

dV×= 2

4π

Substitute the value of r that is 5 into 3s2

and the value ofdt

ds into the formula.

3

3

4rV π= is the formula for the volume of

sphere. Differentiate it that is respect to r.

The value of dt

dVis negative because the rate

is decreasing. If it is increasing, it will be

positive.

Info…



Page | 128

When 13

4−−= scm

dt

dV, cmr 3= ,

1

2

36

4

)3(44

−−=

×=−

cmsdt

dr

dt

dr

π

π

1

9

1 −−= cmsπ

EXERCISE 9.8

1. The area of a circular water ripple expands at the rate of 12

6−

scm when the radius is 5 cm. Hence,

find the rate of change of the radius if the ripple.

2. Given thatx

xy4

5 += . If y increases at a constant rate of 3 unit per second, find the rate of change of

x when x= 4.

9.9 SMALL CHANGE AND APPROXIMATION

1. We have learned that delta x( xδ ) is the small change in x and delta y( yδ ) is the small change in y.

2. We know that x

y

x δ

δδ 0lim

→=

dx

dy.

2. The formula for small change and approximation is

Where x and y are variables that can be changes depends on the case and situation.

Example 1:

Given the radius of a circle increases from 4cm to 4.01.cm. Find the approximate change in its area.

Solution:

In this case, A is the area (A) and B is the radius (r).

Given that cmr 01.0=δ , cmr 4= and 2

rA π= . We are going to find Aδ .

rdr

dA

rA

π

π

2

2

=

=

dx

dy

x

y≈

δ

δ

Substitute the value of r that is 3 into

24 rπ and the value of

dt

dV into the

formula.

2rA π= is the formula for the area of circle.

Differentiate it that is respect to r.



Page | 129

The formula is

dr

dA

r

A≈

δ

δ

Substitute rdr

dAπ2= into the formula,

rr

Aπ

δ

δ2≈

rrA δπδ ×≈ 2

When cmr 01.0=δ , cmr 4= ,

)01.0()4(2 ×≈ πδA

)01.0(8 ×≈ π

2

08.0 cmπ≈

Example 2:

Given 432 += xy . Find the approximate change in y when x increases from 2 to 2.03.

Solution:

In this case, A is y and B is x.

Given that 03.0=xδ , 2=x and 432 += xy . We are going to find yδ .

xdx

dy

xy

6

432

=

+=

The formula is

dx

dy

x

y≈

δ

δ

Substitute xdx

dy6= into the formula,

xxy

xx

y

δδ

δ

δ

.6

6

≈

≈

When 03.0=xδ , 2=x ,

03.0.)2(6 ×≈yδ

36.0≈

EXERCISE 9.9

1. Given 57223 +−= xxy , find the value of

dx

dyat the point (3, -4). Hence, find the small change in y,

when x decreases from 2 to 1.97.

2. Find the small change in the area of a circle if its radius increases from 5cm to 5.02cm.

Move rδ to other side.

Substitute the value of r into and the

value of rδ into the formula.

Find dx

dy

Move xδ to other side.

Substitute the value of x into and the

value of xδ into the formula.



Page | 130

CHAPTER REVIEW EXERCISE

1. Evaluate the following limits:

(a) 35

64lim

−

+

∞→ x

x

x

(b) )2

4(lim

2

2 −

−

→ x

x

x

(c) x

x

x −

+

→ 1

1lim

3

2. Given 1432 +−= xxy . Find

dx

dy.

3. Differentiate 2

35

x

x −with respect to x.

4. Given )1)(2()(2 −+= xxxf . Find )3(''f

5. Given5

)23( += xy . Find the gradient of the curve at the point where x= -1.

6. Given the gradient of the normal to the curve 432 +−= xkxy at x= -2 is

13

1− . Find the value of k.

7. Find the equation of the tangent and the equation of normal to the curve 2

3 xy −= at point (2, -1)

8. Given xyP = and 30=+ yx , find the maximum value of P.

9. P Q

X cm

R S

y cm

The diagram above shows a circle inside rectangle PQRS such that the circle is constantly touching the

two sides of the rectangle. Given the perimeter of PQRS is 60 cm.

(a) Show that the area of shaded region 2

4

430 xxA

+−=

π

(b) Using 142.3=π , find the length and width of the rectangle that make the area of the shaded

region a maximum.

10. Given .57223 +−= xxy Find the rate of change in y, at the instant when x= 3 and the rate of

change ion x is 5 units per second.

11. Given23

2

9)1( ttp +−= . Find

dt

dpand hence find the values of t where 9=

dt

dp.



Page | 131

12. Given that graph of function 2

3)(

x

qpxxf += has a gradient function

3

2 1926)(

xxxf −= where p

and q are constants, find

(a) the values of p and q

(b) the x-coordinate of the turning point of the graph of the function.

13. The straight line kxy =+4 is the normal to the curve 5)32(2 −−= xy at point E. Find

(a) the coordinates of point E and the value of k

(b) the equation of tangent at point E

14. Differentiate the following expressions with respect to x.

(a) 32

)51( x+

(b) 2

434 +

−

x

x

15. Given x

xxf1

5)(3 −= , find )('' xf .

16. Given that xxy 322 −= and xp −= 5 .

(a) Find dp

dywhen x= 2,

(b) Find the small change in x when p increases from 4 to 4.05.

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