DIFFERENTIATIONS FORM 4
MODULE 9(A) ADDITIONAL MATHEMATICS TOPIC : DIFFERENTIATIONS DIFFERENTIATIONS Express Note 1. Limit dy = dx
lim
δy
δx →∞
δx
2. First derivative derivative for polynomial functions (i) Using formula dy n −1 y = axn , @ = anx dx f(x) = axn , f ' (x) = anxn-1
(iv) y = u + v dy du = dx dx
(ii) y = u - v dy du = dx dx
(v) Product of two two polynomial polynomial dy dv du =u + v y = uv , dx dx dx
−
dv dx
(iii) Quotient of two polynomial du dv − u v u dy dx dx , y = = 2 v dx v
+
dv dx
(vi) Composite function dy n −1 y = [f ( x )]n , = n[f ( x )] f ' ( x ) dx
3. Second derivatives of function d 2 y : , f '' (x) dx 2 4. Application / use use of differentiation 1.
dy = gradient of a curve / tangent dx
i) Equation of tangent : dy y − y1 = ( x − x1 ) dx ii) Equation of normal : 1 y − y1 = − ( x − x1 ) dy dx 2. Maximum/Minimum dy d 2 y i) Find , dx dx 2 dy ii) Let = 0, find the values x & y dx
iii) Substitute the value of x in
d 2 y dx 2
If
d 2 y d 2 y > 0, the point is minimum If dx 2 dx 2 < 0, the point is maximum
3. Rate of change change [involve [involve time] dy dy dt = × dx dt dx 4. Small changes & approximation approximat ion dy δ y = δ x dx Approximate value of y = y + δ y
PROGRAM KECEMERLANGAN AKADEMIK (szk smkpn)
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DIFFERENTIATIONS FORM 4
1.
4
Differentiate 3 x 2 ( 2 x − 5 ) with respect to x . [3 marks] [Answer : 6 x(6 x − 5)(2 x − 5)3 ]
2.
3.
2 . Given that y increases x at a constant rate 4 units per second, find the rate of change of x when x = 2. [3 marks] 8 [Answer : ] 5
Two variables, x and y , are related by the equation y
Given that h( x ) =
1 (3 x − 5)2
= 3 x +
, evaluate h ''(1) ''(1) . [4 marks]
[Answer :
4.
5.
6.
27 ] 8
1 3 h + 8h , where h cm is the 3 height of the water in the container. Water is poured into the container at the rate of 10 cm³ s¯¹. Find the rate of change of the height of water, in cm³ s ¯¹, at the instant when its height is 2 cm. [3 marks] 5 [Answer : ] 6
The volume of water, V cm³, in a container is given by V
=
The point P lies on the curve y = ( x − 5)2 . It is given that the gradient of the normal at 1 P is − . Find the coordinates of P . 4 [3 marks] [Answer : (7 : (7 , 4) ] 4) ] It is given that y
=
dy 2 7 in terms of x . u , where u = 3 x − 5 . Find 3 dx [4 marks] 6
[Answer : 14 ( 3 x − 5 ) ] 7.
Given that y
2
= 3x +
x − 4 ,
dy when x = 1, dx (b) express the approximate change in y , in terms of p , when x changes from 1 to 1 + p , where p is a small value. [4 marks] [Answer : a) 7 b) 7p]
(a) find the value of
PROGRAM KECEMERLANGAN AKADEMIK (szk smkpn)
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DIFFERENTIATIONS FORM 4
8.
9.
dy = 3kx + 5 , where k is a constant. The gradient of dx the curve at x = x = 2 is 9. Find the value of k . [2 marks] 2 [Answer : ] 3
The curve y
= f ( x )
The curve y = x 2 the value of p .
is such that
− 32 x + 6 4
has a minimum point at x = x = p , where p is p is a constant. Find [3 marks]
[Answer : 16 ] 10.
4
Differentiate 2 x 3 ( 3 x − 5 ) with respect to x. [3 marks] [Answer : 6 x 2 (3 x − 5 )3 (7 x − 5 ) ]
11.
Find the rate of change of the area of the circle ifif the rate of change of the radius is 0·2 cms – 1 when the radius is 4 cm. [3 marks] [Answer : 1.6π ]
12.
Given that y = y = 4x 4x ((x – x – 2), calculate (a) the value of x when x when y is y is minimum, (b) the minimum value of y . [4 marks] [Answer : a) 1 b) -4 ]
13.
A point R lies R lies on the curve y = y = 4x 2 – 2x + x + 3. Given that the gradient of the normal to 1 the curve at R is R is − , find the coordinates of R . 6 [3 marks] [Answer : (1 , 5) ]
14.
Given that y = (2x – – 3) 3) – 5, 5, fin find d y = (2x dy (a) the value of at the point (2, −4), dx (b) the small change in y if y if x decreases from 2 to 1 ⋅97. [4 marks] [Answer : a) 4 b) -0.12 ]
15.
Given that the curve y = y =
6 , find 2 x − 1
(a) the gradient at the point (−1, −2), (b) the equation of the normal to the curve at that point. [4 marks] [Answer : a) −
4 3 5 b) y = x − ] 3 4 4
PROGRAM KECEMERLANGAN AKADEMIK (szk smkpn)
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DIFFERENTIATIONS
16.
17.
FORM 4
Find the small change of the surface area of the cube when the sides increase from 4 cm to 4.02 cm. [3 marks] [Answer : 0.96 ] Given that y = y = (1 + 2x 2x )5, find the value of
dy when x = x = −1. dx [2 marks]
[Answer : 10 ] 18
19.
The volume of the liquid in a container, V cm V cm3 is given by V = V = 2x 3 + 4x 2 + 5, where x cm is the depth of the liquid in the container. Given that V increases at a rate of 32 cm3s−1, find the rate of increase of x when x when x = x = 2. [4 marks] 4 [Answer : ] 5 Given f ( x ) =
2 , find the value of f ′(2) . (5 − 3 x )3 [3 marks]
[Answer : 18 ]
20.
The radius of circle decreases at the rate of 0.5 cm¯¹. Find the rate of change of the area when the radius is 4 cm. [3 marks] [Answer : −4π ]
21.
Find the equation of the normal to the curve y = 3 x 2 [Answer : 2y
22.
=
− 8 x + 1
at the point (1, −4). [3 marks]
x − 9 ]
Find the value of
d 3x 2 ( dx
3
− 4 x − 6
) when x = 2. [2 marks]
[Answer : -36 ]
23.
Find the gradient of the normal to the curve y
= ( 2 x + 3 )
3
when x = -1 [3 marks]
1 [Answer : − ] 6 24.
25.
The radius of of a sphere increases from 4 cm to 8 cm in two seconds. Find the rate of increase of the surface area of the sphere. [4 marks] [Answer : 64π ] It is given that y
=
3 6 u 5
+
4 , where u
= 5 x − 3 ,
find
dy in terms of x . dx
[3 marks] 5
[Answer : 18 ( 5 x − 3 ) ]
PROGRAM KECEMERLANGAN AKADEMIK (szk smkpn)
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DIFFERENTIATIONS FORM 4
26.
Find the gradient of the normal to the curve y
= ( 2 x + 3 )
3
when x = -1. [3 marks]
1 [Answer : − ] 6 27.
Two variables, x and y , are related by the equation y
=
2
x ( 3 − x ) . Given that y
increases at constant rate of of 4 unit per second, second, find the rate of of change of x when x = 2. [3 marks] 4 [Answer : − ] 3
28.
The curve y = 2ax 2 + bx has a turning point at (2 , 3), where a and b are constants. Find the valueof a and of b . [3 marks] 3 [Answer : a= − b= 3 ] 8
29.
Given that g (x ) = x 2 (2 – x )4, (a) find g’ (x) (b) evaluate g’(1) . [4 marks] [Answer : a) 2x ( 2 − x )
30.
3
( 2 − 3 x ) b) -2 ]
n 2 + n − 2 Without constructing a table, find the value of lim . n →1 − 1 n [2 marks] [Answer : 3 ]
31.
Given y =
k dy = g ( x ) , find the value of k if and 3 dx ( 2x − 5)
∫
3
2
[ g ( x ) + 1] dx = 7 . [3 marks]
[Answer : 3 ]
32.
33.
2 . Given q increases at a q constant rate of 4 units per second when p = 2, find the rate of change in p . [3 marks] [Answer : 10 ]
Two variables p and q are related by the equation p = 3q +
Given
f (x ) = (5 - 3x )4, find f ’’(2).
[3 marks] [Answer : 108 ]
PROGRAM KECEMERLANGAN AKADEMIK (szk smkpn)
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DIFFERENTIATIONS FORM 4
34.
35.
2 3 π r , r is the radius. Find the 3 approximate change in V if r increase from 3 cm to 3.005 cm. Give your answer in term of π . [4 marks] [Answer : 0.33π ]
Volume, V cm³, of a solid is given by V
= 8π r
2
+
x 2 − 3 x Find the value of lim 2 x → 2 x − 4 x + 3 [2 marks] [Answer : 2 ]
36.
Given that f ( x ) =
3 x + 4 , find f '(2) . 3 − x 2 [3 marks]
[Answer : 37 ]
37.
It is given that y
2
= ( 3 x −
4)5 , evaluate
dy , when x = 1. dx
[3 marks] [Answer : 30 ]
38.
Find the coordinates of the stationary points on the curve y
=
x3
2
− 6 x + 17 .
[3 marks] [Answer : (0,17), (4,-15) ]
39.
The side of a cube changes at the rate of 0.3 cm s -1. Find the rate of change of of its volume when its surface area is 216 cm 2. [4 marks] [Answer : 32.4 ]
40. Given that f ( x ) =
( 2x − 1) x − 1
3
, x ≠ 1 , find f’(x). [3 marks]
2
[Answer :
( 2 x − 1) ( 4 x − 5 ) ] 2 − 1 ( x )
END OF MODULE
PROGRAM KECEMERLANGAN AKADEMIK (szk smkpn)
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DIFFERENTIATIONS FORM 4
ANSWERS MODULE 9(A) ADDITIONAL MATHEMATICS TOPIC : DIFFERENTIATION DIFFERENTIATION 1.
y
= 3x
2
(2x − 5)4
dy 2 3 ( 2)) + (2 x − 5)4 (6 x ) = 3 x ( 4( 2 x − 5) (2 dx
2.
y
=
6 x(2 ( 2 x − 5)3 [ 4 x + (2 x − 5)]
=
6 x(6 x − 5)(2 x − 5)3
= 3 x +
2 x 2
dy =3− dx x 2 dy dy dx = × dt dx dt 2 dx 4 = 3 − 2 × x dt
4 = 3 −
2 dx × (2)2 dt
dx 8 = dt 5
3.
h( x ) = (3 x − 5)−2 h '( x ) = −6(3 x − 5)−3 h ''( x ) = 54(3 x − 5)−4 h ''(1) = 54( −2)−4 =
4.
V
=
27 8
1 3 h 3
+ 8h
dV 2 = h + 8 dh dV dV dh = × dt d h dt dh 10 = ( h 2 + 8) × dt dh 10 = = 0.8333 dt 22 + 8
PROGRAM KECEMERLANGAN AKADEMIK (szk smkpn)
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DIFFERENTIATIONS FORM 4
5.
y
= ( x − 5)
2
dy = 2( x − 5) dx 1 − m 2 = −1 4 m 2 = 4
2( x − 5) = 4 x = 7 y = (7 − 5)2
=
4
P (7,4)
6.
2 7 2 u = (3 x − 5)7 3 3 dy 2 6 = (7)(3 x − 5) (3 ) dx 3
y
=
6
= 14(3 x − 5)
7.
(a)
y
= 3x
2
+
x − 4
dy = 6 x + 1 dx dy = 6(1) + 1 = 7 x = 1, dx
(b)
δ x ≈
p
δ y
dy dx
≈
δ x δ y
p
≈7
δ y ≈
8.
3k (2 ( 2) + 5 = 9 k =
9.
7 p
y
=
2 3 x2
− 32 x + 64
dy = 2 x − 32 dx min x = p, 2( p ) − 32 = 0 p = 16
PROGRAM KECEMERLANGAN AKADEMIK (szk smkpn)
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DIFFERENTIATIONS FORM 4
10.
y
= 2x
3
(3 x − 5)4
dy dv du =u + v dx dx dx dy 3 3 = 2 x (12)(3 x − 5) dx = 6x
11.
2
+ (3 x − 5 )
4
(6 (6 x 2 )
(3 x − 5)3 (7 x − 5)
A = π r 2 dA dr dA dt dA dt
= 2π r =
dA dr × dr dt
= 2π ( 4) × 0.2 = 1.6π
12.
a)
dy = 8 x − 8 = 0 dx x = 1
b)
x = 1 =
4(1)2
− 8(1)
= −4
13.
m1m2
= −1
, m = 6
dy = 8 x − 2 = 6 dx x = 1 y = 4(1)2
− 2(1) + 3 = 5
R (1,5)
14.
a)
dy = 4( 2 x − 3) dx x = 2 dy = 4( 2( 2) − 3) dx =4
b)
δ x = −0.03
dy × δ x dx = 4 × −0.03
δy =
= −0.12
PROGRAM KECEMERLANGAN AKADEMIK (szk smkpn)
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DIFFERENTIATIONS FORM 4
15.
a)
dy −2 = −12 ( 2 x − 1) dx −12 =
( 2 x − 1)
2
x = −1 −12 dy = dx (2(1) − 1)2
b)
16
m =
=−
4 3
3 4 3 ( x + 1) 4 3 5 x − 4 4
y
+2=
y
=
A = 6 x 2 dA = 12 x dx dA × δ x δl = dx = 12( 4) × 0.02 = 0.96
17.
dy 4 = 10 (1 + 2 x ) dx 4 = 10(1 + 2( −1)) = 10
18.
dV 2 = 6 x + 8 x dx dx dx dV = × dt dV dt 1 = × 32 6(2)2 + 8(2) =
19.
4 5
f '( x ) = f '(2) =
18
( 5 − 3 x )
4
18
( 5 − 3(2) )
4
= 18
PROGRAM KECEMERLANGAN AKADEMIK (szk smkpn)
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DIFFERENTIATIONS FORM 4
20.
A = π r 2 dA dr dA dt
= 2π r
dA dr × dr dt = 2π ( 4 ) × ( −0.5 ) =
= −4π
21.
dy = 6 x − 8 dx m1 = −2 m 2
=
1 2
1 ( x − 1) 2 2y = x − 9 y
22.
+4=
dy 2 = 6 x − 12 x dx 2 = 6( 2) − 12( 2) = −36
23.
2 dy = 6 ( 2 x + 3 ) dx
=
6 ( 2 ( −1) + 3 )
=
6
Normal = − 24.
2
1 6
A = 4π r 2 dA dr dA dt
dr =2 dt dA dr = × dr dt = 8π ( 4 ) × 2 = 8π r
= 64π
25.
dy du dy dx
18 5 u 5 dy du = × du dx 18 5 = u × 5 5
du dx
=
= 18 ( 5 x − 3 )
=
5
5
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DIFFERENTIATIONS FORM 4
26.
dy 2 = 6 ( 2 x + 3 ) dx = 6 ( 2( −1) + 3 )
2
=6
m = −
27.
y
1 6
= 9x − 6x
2
+
x3
dy = 9 − 12 1 2 x + 2 x 2 dx 2
= 9 − 12( 2) + 3( 2) = −3
dy dx
dt dx dt −3 = 4 × dx 4 dx =− dt 3
28.
=
dy dt
×
3 = 2a(2)2
dy = 4ax + b = 0 dx 4a(2) + b = 0
+ 2b
8a + 2b = 3
8a + b = 0 a=−
29.
3 , b = 3 8
a)
dy 4 3 2 = ( 2 − x ) ( 2 x ) + x ( 4 ) ( 2 − x ) ( −1) dx = 2x ( 2 −
b)
=
x)
2 (1) ( 2 − (1) )
3
3
( 2 − 3 x )
( 2 − 3 (1) )
= −2
30.
=
=
n2
+ n − 2
n − 1 ( n − 1) ( n + 2 )
n − 1 = ( n + 2 ) = 1+ 2 =
3
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DIFFERENTIATIONS FORM 4
31.
∫
3
2
g ( x )dx
∫
+
3
2
dx = 7
3
k 3 (2x − 5)3 + [ x ]2 = 7 2 k k − + [3 − 2] = 7 3 3 − − ( 2 3 5 ) ( 2 2 5 ) ( ) ( ) k = 3
32.
2 dp =3− 2 dq q dp dt
=
dp d q × dq d t
2 dp = 3 − 2 × 4 dt 2 = 10
33.
f '( x ) = −12 ( 5 − 3 x )
3
f ''( x ) = 108 ( 5 − 3 x )
2
f ''(2) = 108 ( 5 − 3(2) )
2
= 108
34.
dv 2 = 16π r + 2π r dr dv × δ x δy = dr δ y =
16π ( 3 ) + 2π ( 3 )2 × 0.005
= 0.33π
35.
x 2 − 3 x = lim x → 2 x 2 − 4 x + 3 x ( x − 3 x ) = lim x → 2 ( x − 1) ( x − 3 x ) x = lim x → 2 x − 1 2 = lim =2 x → 2 2 − 1
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DIFFERENTIATIONS FORM 4
36.
f ′( x ) =
(3 − x 2 )(3) − (3 x + 4)( −2 x ) (3 − x 2 )2
f ′(2) =
3 − (2)2 (3) − [(3(2) + 4)( −2(2))] 2 3 − (2)2
= 37
37.
dy 2 = 5 3x dx x = 1
(
=5
3 12
( )
−4
)
4
( 6 x )
4
−4
6 (1)
= 30
38.
dy 2 = 3 x − 12 x dx 3 x 2 − 12 x = 0
3 x ( x − 4 ) = 0 x
=0
, x = 4
x
= 0,y = 17
x
=
4, y
= −15
(0,17) (4, (4, −15) 39.
v
=
x 3
dv 2 = 3 x dx 3
216
δ v =
=
6
3 ( 6)
2
× 0 .3
= 32.4
40.
(
2
) − ( 2x − 1)
3
dy ( x − 1) 3 ( 2 x − 1) ( 2) = 2 dx ( x − 1)
(1)
2
=
( 2x − 1) 6 ( x − 1) − ( 2 x − 1) ( x − 1)
2
2
( 2x − 1) ( 4 x − 5 ) = 2 ( x − 1)
END OF MODULE
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