Page 30 - Ranger SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation

Example 8 (b) Let u = (x + 1) and v = (x + 3) 1 2
2
dy du
Find for each of the following. Then, = 2(x + 1) and
dx dx
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1
(a) y = x – 2 dv = (x + 3) – 1 2
3 – x 2 dx 2
(x + 1) 2
(b) y = = 1
x
 + 3 2x + 3
Solution v du – u dv
(a) Let u = x – 2 and v = 3 – x 2 dy = dx 2 dx
du dx v
Then, = 1 and 2
dx  + 3(2)(x + 1) – (x + 1)
x
dv 2 + 3
x
dx = –2x =  + 3 2
x
By using the quotient rule, 2(x + 3)(2)(x + 1) – (x + 1) 2
du dv =
x
dy v dx – u dx 2(x + 3) + 3
=
dx v 2 (x + 1)(4x + 12 – x – 1)
(3 – x )(1) – (x – 2)(–2x) =
2
= 2(x + 3)x + 3
(3 – x )
2 2
3 – x + 2x – 4x = (x + 1)(3x + 11)
2
2
= 2(x + 3)x + 3
(3 – x )
2 2
x – 4x + 3
2
= (3 – x )
2 2
Example 9
b
6
Alternative Method Given y = ax + and dy = 2x – x n – 1 ,
n
4
x
dx
x – 2 find the values of a, b and n.
y = = (x – 2)(3 – x )
2 –1
3 – x 2
Let u = x – 2 and v = (3 – x ) Solution
2 –1
dv
du = 1 and = –1(3 – x ) (–2x)
2 –2
4
dx dx y = ax + bx –1
2x
2 2
= (3 – x ) dy = 4ax – bx –2
3
By using the multiplication rule dx b
3
dy = (x – 2) 2x 2 2 + 1 = 4ax – x 2
dx  (3 – x ) 3 – x 2
2
= 2x(x – 2) + (3 – x ) By comparing, Form 5
(3 – x ) 4a = 2
2 2
2x – 4x + 3 – x 2 1
2
= a =
(3 – x ) 2
2 2
x – 4x + 3 b = 6
2
=
(3 – x ) n = 3
2 2
183
02 Ranger Mate Tambahan Tg5.indd 183 25/02/2022 9:23 AM