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8/12/2019 Addmath Differentiation
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
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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
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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
Gradient of tangent, mT
=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
Gradient of normal,4
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
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8/12/2019 Addmath Differentiation
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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
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Topic : DIFFERENTIATION
Unit J : Problems of Rates of Change
Task 1 : Answer all the questions below.
(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
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Task 2 : Answer all the questions below.
(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 )
Answer :
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
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Task 3 : Answer all the questions below.
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
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(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
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(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
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Topic : DIFFERENTIATION
Unit K : Problems of Small Changes and Approximations
Task 1 : Answer all the questions below.
(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
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Task 2 : Answer all the questions below.
(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
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Task 3 : Answer all the questions below.
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
approximate increase in its radius.
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
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Task 4 : Answer all the questions below.
(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
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Task 5 : Answer all the questions below.
(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
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Topic : DIFFERENTIATION
Unit L : Problems of Maximum and Minimum Values of a Function.
Task 1 : Answer all questions below.
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
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