Differentiation Guide for Additional Mathematics Module Form 4

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(lim
2
+
→
x
x
= 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(lim
2
=+
→
x
x
4. Hence, the function 2+= xy has limit 4 as 2→x .
Example 2:
2
4
lim
2
2 −
−
→ x
x
x
)2(
)2)(2(
lim
2 −
−+
→ x
xx
x
)2(lim
2
+
→
x
x
22 +=
4=
2+= xy
First of all, factorize the numerator first.
Then simplify.

Page | 106
Example 3:
)
1
(lim
x
x
x
+
∞→
)
1
(lim
xx
x
x
+=
∞→
)
1
1(lim
xx
+=
∞→
01+=
1=
EXERCISE 9.1
Find:
(a) 2
2
23
lim
x
x
x
+
∞→
(b) )32(lim
1
+
→
x
x
(c) 2
5
2lim x
x→
(d)
3
9
lim
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
0
1
=
∞
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
1
and becomes
∞
1
. We know that
0
1
=
∞
.
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
y
mPQ
δ
δ
= .
5. But at point P,
x
y
m
x
P
δ
δ
δ 0
lim
→
=
6.
x
y
x δ
δ
δ 0
lim
→
is the gradient of tangent at point P.
7. To write
x
y
x δ
δ
δ 0
lim
→
, it is a quite long, so we can write them as
dx
dy
.
8.
dx
dy
is differentiation.
9.22 DIFFERENTIATION BY THE FIRST PRINCIPLE
Q( x+ xδ , y + yδ )
P(x, y)
2
xy =
22
2
2
)(
xxxxyy
xxyy
δδδ
δδ
++=+
+=+
2
xy =
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
- ,
xx
x
y
xx
x
y
x
xxx
x
y
xxxy
xxxxxyyy
xx
δ
δ
δ
δ
δ
δ
δ
δδ
δ
δ
δδδ
δδδ
δδ
+=
+=
+
=
+=
−++=−+
→→
2limlim
2
2
2
2
00
2
2
222
x
dx
dy
2=
EXERCISE 9.20
Find
dx
dy
of the following equation using the first principle.
(a) 2
3xy = (b) 52 2
+= xy
(c)
x
y
2
= (d)
x
y
3
5 −=
(e) 2
31 xy −= (f) xxy 53 2
−=
- Finding
dx
dy
by using formula -
n
axy =
Given that y = f(x),
The first derivate
dx
dy
is equivalent to )('
xf ,
That is if n
axxf =)( ,
Then,
)(
)( '
xf
dx
xdf
=
2 1
1
.. −
= n
xna
dx
dy
Left hand side and right hand side are
divided by xδ to make the term
x
y
δ
δ
. The
right hand side is then simplified.
For left hand side, we already know that
x
y
x δ
δ
δ 0
lim
→
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
..)(' −
= n
xnaxf
If the 2
xy = , then x
dx
dy
2= .
If 2
)( xxf = , then xxf 2)('
=
If xy 2=
2
)2()(
=
=
dx
dy
x
dx
d
y
dx
d
If xxf 2)( =
2)('
2
)(
)2()]([
=
=
=
xf
dx
xdf
x
dx
d
xf
dx
d

Page | 109
Example 1:
2
xy =
)( 2
x
dx
d
dx
dy
=
12
.2.1 −
= x
dx
dy
x
dx
dy
2=
Example 2:
xxf
xxxf
xxf
6)('
.0.6.2.3)('
63)(
1012
2
=
+=
+=
−−
EXERCISE 9.21
1. Find
dx
dy
for each of the following equation. Hence, find the value of
dx
dy
at point where 2=x .
(a) 42 4
−= xy
(b) )5( 2
+= xxy
(c) 2
2
2
x
xx
y
+
=
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 4
2
1 4
−x with respect to x.
Solution:
)4
2
1
( 04
xx
dx
d
−
1014
4.0
2
1
.4 −−
−= xx
3
2x=
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
)4
2
1
( 4
−x
dx
d
, cannot with 4
2
1 4
−x .

Page | 110
Example 2:
Differentiate )32)(32( +− xx with respect to x.
x
xx
xx
dx
d
xx
dx
d
8
.9.0.4.2
)94(
)]32)(32[(
1012
02
=
−=
−
+−
−−
EXERCISE 9.22
Differentiate each of the following with respect to x.
(a) 53 2
−x
(b)
x
x
22
−
(c) )43)(43( +− xx
9.3 FIRST DERIVATIVE OF COMPOSITE FUNCTION
Example 1:
Find
dx
dy
for the function 2
)12( += xy
Solution:
Method 1
2
)12( += xy
144 2
++= xxy
Method 2
2
)12( += xy
Let
12 += xu
2=
dx
du
2
uy =
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 += x
dx
dy
We know the value
du
dy
and
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
u
du
dy
2=
22 ×= u
dx
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.
2
uy =
u
du
dy
2=
2. Then, differentiate the part in the bracket.
12 += xu
2=
dx
du
3. Multiply both of them and it will result
dx
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
dy
for the function 5
)43( += xy
Let
43 += xu
3=
dx
du
5
uy =
4
5u
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
35 4
×= u
dx
dy
4
15u=
4
)43(15 += x
We can also write directly like this:
3.)43(5 15−
+= x
dx
dy
4
)43(15 += x
EXERCISE 9.3
Find
dx
dy
for each of the following equation.
(a) 4
)31( xy −=
(b) 3
)
1
1(
x
y +=
(c) 4
)53(
1
−
=
x
y
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
- , ,
)(limlim
00 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
v
u
x
u
v
x
v
u
x
y
dxdxdxdxdxdxdx δ
δ
δ
δ
δ
δ
δ
δ
δ
0000000
limlimlimlimlimlimlim
→→→→→→→
×××××=
dx
dv
dx
du
v
dx
dv
u
dx
dy
.0++=
Example:
Find
dx
dy
for the function )1)(12( ++= xxy
)1)(12( ++= xxy
u v
1
1
2
12
=
+=
=
+=
dx
dv
xv
dx
du
xu
The formula is
dx
du
v
dx
dv
u
dx
dy
+= , just substitute each term into the formula to find
dx
dy
)2)(1()1)(12( +++= xx
dx
dy
34
2212
+=
+++=
x
xx
2 1
dx
du
v
dx
dv
u
dx
dy
+=

Page | 114
EXERCISE 9.40
1. Find
dx
dy
dx
dy
at 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
−=
−=
=
=
x
dx
dv
xv
x
dx
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
xx
dx
d
If it is given an equation, the formula used is
dx
du
v
dx
dv
u
dx
dy
+= . If it is given an expression not
equation, the formula used is just
dx
du
v
dx
dv
u + 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
x
y
v
u
y =
vv
uu
yy
δ
δ
δ
+
+
=+
- , ,












+
−
=
+
−
=
×
+
−
=
+
−
=
+
−−+
=
+
+
−
+
+
=
−
+
+
=−+
→→ )(
limlim
)(
1
)(
)(
)(
)(
)(
)(
)(
00 vvv
x
v
u
x
u
v
x
y
vvv
x
v
u
x
u
v
x
y
xvvv
vuuv
x
y
vvv
vuuv
y
vvv
vuuvuvuv
y
vvv
vvu
vvv
uuv
y
v
u
vv
uu
yyy
dxdx δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δδ
δδ
δ
δ
δ
δδ
δ
δ
δδ
δ
δ
δ
δ
δ
δ
δ
δ
δ
u
v
2
2 1
1 How to get the formula?

Page | 116
)(limlim
limlimlimlim
)(
lim
00
0000
0
vvv
x
v
u
x
u
v
vvv
x
v
u
x
u
v
dxdx
dxdxdxdx
dx
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
+×
×−×
=












+
−
=
→→
→→→→
→
vv
dx
dv
u
dx
du
v
dx
dy
.
−
=
Example:
Find
dx
dy
for the function
1
12
+
+
=
x
x
y by using formula 2
v
dx
dv
u
dx
du
v
dx
dy
−
= .
1
1
2
12
=
+=
=
+=
dx
dv
xv
dx
du
xu
Substitute each term into the formula to find
dx
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
2
v
dx
dv
u
dx
du
v
dx
dy
−
=

Page | 117
EXERCISE 9.42
1. Find
dx
dy
dx
dy
(a)
( )
52
23
2
−
+
=
x
x
y
(b)
( )
( )2
3
1
12
+
+
=
x
x
y
(c)
( )
( )3
2
74
65
−
−
=
x
x
y
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
xx
xf
(b)
( )
52
23
3
−
+
=
x
x
y
(c)
( )
xx
x
y
52
23
2
2
−
+
=
Example:
Differentiate
1
23 2
+
+
x
xx
with respect to x.
1
1
26
23 2
=
+=
+=
+=
dx
dv
xv
x
dx
du
xxu






+
+
1
23 2
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
2
v
dx
dv
u
dx
du
v
dx
dy
−
= . If it is given an expression not
equation, the formula used is just
2
v
dx
dv
u
dx
du
v −
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
3
4
x
xx +
(c)
12
5 2
+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
dy
is 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 592 2
−−= xxy at the point
(3, 4)
Solution:
94
592 2
−=
−−=
x
dx
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
3
3
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 243 2
+−= xxy that is parallel to the line 52 =− xy .
Solution:
46
243 2
−=
+−=
x
dx
dy
xxy
2
5
2
1
52
+=
=−
xy
xy
m =
2
1
normalm is parallel to this line, so
2
1
=normalm .
2
1
2
1
1
tan
tan
tan
−=
−=×
−=×
gent
gent
normalgent
m
m
mm
We know that 46 −= x
dx
dy
and 2tan −=gentm , hence
3
1
246
=
−=−
x
x
2tan −=gentm and
2
1
=normalm at point which
3
1
=x .
243 2
+−= xxy
3
1
=x ,
2
3
1
4
3
1
3
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 −= x
dx
dy
that 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
1
respectively. We substitute the value of
x into the equation of the curve that is
243 2
+−= 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
1
22
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 46 23
+−= ; 3=x
(b) 2
)3)(2( −+= xxy ; 2−=x
(c)
23
32
−
+
=
x
x
y ; 1=x
9.6 SECOND ORDER DIFFERENTIATION
1. First differentiation is
dx
dy
.
2. Second order differentiation is 2
2
dx
yd
that is we differentiate for the second times.
Example 1:
Given xxxy 46 23
+−= ,find 2
2
dx
yd
.
126
4123
46
2
2
2
23
−=
+−=
+−=
x
dx
yd
xx
dx
dy
xxxy
Multiply the equation by 3
Differentiate one more time
dx
d






dx
dy
= 2
2
dx
yd

Page | 122
Example 1:
Given
x
xxxf
1
4)( 3
++= ,find )(''
xf .
3
''
3''
22'
13
3
2
6)(
26)(
143)(
4)(
1
4)(
x
xxf
xxxf
xxxf
xxxxf
x
xxxf
+=
+=
−+=
++=
++=
−
−
−
EXERCISE 9.6
1. Find 2
2
dx
yd
for each of the following equation. Hence, find the value of 2
2
dx
yd
(a) xxy 52 4
−=
(b) )5(2 2
xxxy +=
(c) 2
2
32
x
xx
y
+
=
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
xf
dx
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
dy
is 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 593 23
+−−= xxxy . Hence, state each point is minimum or
maximum.
Solution:
Method 1
963
593
2
23
−−=
+−−=
xx
dx
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(3 2
=
−−−−=
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(3 2
=
−−=
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(3 2
−=
−−=
dx
dy
dx
dy
15
9)4(6)4(3 2
=
−−=
dx
dy
dx
dy
Factorize to find the values of x
Substitute the value of x into the equation of the curve
that is 593 23
+−−= 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
−−=
+−−=
xx
dx
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).
963 2
−−= xx
dx
dy
662
2
−= x
dx
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 593 23
+−−= 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 02
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
xy
1
+=
(b) 2
2 16
)(
x
xxf +=
2. Find the turning point at the following curve
(a ) 2
8 xxy −=
(b ) 2
2 96
2)(
x
xxf +=
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 3
sV = . We are going to find
dt
dV
.
2
3
3s
ds
dV
sV
=
=
dt
dB
dB
dA
dt
dA
×=
3
sV = 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
×=
Substitute 2
3s
ds
dV
= into the formula.
dt
ds
s
dt
dV
×= )3( 2
When 1
4.1 −
= cms
dt
ds
, 5=s ,
4.1)5(3 2
×=
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
4
rV π= . We are going to find
dt
dr
.
2
3
4
3
4
r
dr
dV
rV
π
π
=
=
The formula is
dt
dr
dr
dV
dt
dV
×=
Substitute 2
4 r
dr
dV
π= into the formula,
dt
dr
r
dt
dV
×= 2
4π
Substitute the value of r that is 5 into 3s2
and the value of
dt
ds
into the formula.
3
3
4
rV π= is the formula for the volume of
sphere. Differentiate it that is respect to r.
The value of
dt
dV
is 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
−−
=
×=−
cms
dt
dr
dt
dr
π
π
1
9
1 −
−= cms
π
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 that
x
xy
4
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
δ
δ 0
lim
→
=
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 .
r
dr
dA
rA
π
π
2
2
=
=
dx
dy
x
y
≈
δ
δ
Substitute the value of r that is 3 into
2
4 rπ and the value of
dt
dV
into the
formula.
2
rA 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 r
dr
dA
π2= into the formula,
r
r
A
π
δ
δ
2≈
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 43 2
+= 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 43 2
+= xy . We are going to find yδ .
x
dx
dy
xy
6
43 2
=
+=
The formula is
dx
dy
x
y
≈
δ
δ
Substitute x
dx
dy
6= into the formula,
xxy
x
x
y
δδ
δ
δ
.6
6
≈
≈
When 03.0=xδ , 2=x ,
03.0.)2(6 ×≈yδ
36.0≈
EXERCISE 9.9
1. Given 572 23
+−= xxy , find the value of
dx
dy
at 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
64
lim
−
+
∞→ x
x
x
(b) )
2
4
(lim
2
2 −
−
→ x
x
x
(c)
x
x
x −
+
→ 1
1
lim
3
2. Given 143 2
+−= xxy . Find
dx
dy
.
3. Differentiate 2
3
5
x
x −
with respect to x.
4. Given )1)(2()( 2
−+= xxxf . Find )3(''f
5. Given 5
)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
4
30 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 .572 23
+−= 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. Given 23
2
9
)1( ttp +−= . Find
dt
dp
and hence find the values of t where 9=
dt
dp
.

Page | 131
12. Given that graph of function 2
3
)(
x
q
pxxf += has a gradient function 3
2 192
6)(
x
xxf −= 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
43
4
+
−
x
x
15. Given
x
xxf
1
5)( 3
−= , find )('' xf .
16. Given that xxy 32 2
−= and xp −= 5 .
(a) Find
dp
dy
when x= 2,
(b) Find the small change in x when p increases from 4 to 4.05.

Differentiation Guide for Additional Mathematics Module Form 4

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