Page 39 - Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation
4. This formula can be verified by using the idea of From the product formula derived,
limit as follows: dy = u dv + v du
u
Let y = …… 1 dx dx dx
v = (3x)[6(2x + 1) ] + (2x + 1) (3)
3
2
From the idea of limit, = 18x(2x + 1) + 3(2x + 1)
2
3
y + dy = u + du …… 2 = 3(2x + 1) (6x + 2x + 1)
2
v + dv = 3(2x + 1) (8x + 1)
2
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Substitute 1 into 2,
u + dy = u + du (b) f(x) = x + 3
v v + dv 3 – 2x
dy = u + du – u Let u = x + 3 and v = 3 – 2x
v + dv v du dv
dy = v(u + du) – u(v + dv) dx = 1 dx = –2
(v + dv)(v)
dy = uv + vdu – uv – udv From quotient formula derived,
(v + dv)(v) v du – u dv
dy = vdu – udv f'(x) = dx 2 dx
v(v + dv) v
Divide both sides with dx, = (3 – 2x)(1) – (x + 3)(–2)
2
(3 – 2x)
dy = vdu – udv ÷ dx
dx v(v + dv) = 3 – 2x + 2x + 6
2
v du – u dv (3 – 2x)
9
= dx dx =
v(v + dv) (3 – 2x) 2
Through limit, Try Questions 10 – 11 in ‘Try This! 2.2’
v lim du – u lim dv
lim dy = dx → 0 dx dx → 0 dx
dx → 0 dx v lim (v + dv) Try This! 2.2
dx → 0
It is known that when dx → 0, dv → 0 and 1. Differentiate each of the following with respect to x.
lim (v + dv) = v. 7
6
dx → 0 (a) y = 5x (b) y = 4x 2
Therefore, (c) y = √x (d) y = 6x 3 2
v lim du – u lim dv 4
lim dy = dx → 0 dx dx → 0 dx (e) y = – 9 (f) y = 3 √x
dx → 0 dx v(v) 3 √x 4

v du – u dv 2. Find f'(x) for each of the following functions.
1 2
and dy = d u = dx dx (a) f(x) = 4x + 5x – 8x
3
2
dx dx v v 2
3
Form 5
(b) f(x) = 3x + 8x 2
5
8 (c) f(x) = + 2 – 3 + 4
1
x
x
x
2
3
Differentiate each of the following function with (d) f(x) = x (1 + 3x – x ) x 4
4
2
respect to x. (e) f(x) = 3√x (1 – 5√x )
(a) y = 3x(2x + 1) (f) f(x) = (3 – √x )
3
2
(b) f(x) = x + 3
3 – 2x 3 dy
Solution 3. Given 12x = 3y, determine the value of dx if
(a) Let u = 3x and v = (2x + 1) (a) x = – 1
3
du = 3 dv = 3(2x + 1) (2) 4 2
2
dx dx = 6(2x + 1) (b) x = 3
2
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