Page 38 - Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation
Solution
(a) Let u = 3x + 2 and y = u REMEMBER!
4
du = 3 dy = 4u
3
dx du Do not forget to differentiate the terms inside the brackets.
\ dy = dy × du
dx du dx Try Questions 5 – 9 in ‘Try This! 2.2’
= 4u × 3
3
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= 12u
3
= 12(3x + 2) D Determining the first derivative of
3
a function involving product and
quotient of algebraic expressions
(b) Let u = 1 – 3x and y = 5u
6
du = –3 dy = 30u 1. The first derivative of the function y = uv such
5
dx du that u = f(x) and v = g(x) is obtained from the
\ dy = dy × du formula
dx du dx
d
= 30u × (–3) dx (uv) = u dv + v du
5
dx
dx
= –90u
5
= –90(1 – 3x)
5
2. This formula can be verified by using the idea of
7 limit as follows:
Let y = uv …… 1
Differentiate each of the following function with respect From the idea of limit,
1 2
to x by using the formula dy = nu n – 1 du . y + dy = (u + du)(v + dv)
dx dx y + dy = uv + udv + vdu + dudv …… 2
(a) y = 4 (b) y = √1 – 5x
2x – 1 Substitute 1 into 2,
uv + dy = uv + udv + vdu + dudv
Solution dy = udv + vdu + dudv
(a) y = 4
2x – 1 Divide each term with dx,
–1
y = 4(2x – 1) dy = u dv + v du + dudv
dy = –1[4(2x – 1) –1 – 1 ] 3 d (2x – 1) dx dx dx dx
4
dx dx Through limit,
= –4(2x – 1) (2)
–2
lim du
lim dudv
lim dy
lim dv
= –8(2x – 1) dx → 0 dx = u dx → 0 dx + v dx → 0 dx + dx → 0 dx
–2
= – 8
(2x – 1) 2 It is known that when dx → 0, dudv → 0.
SPM Tips Therefore,
lim dy = u lim dv + v lim du Form 5
n
–1
Assume 4(2x – 1) as 4x and differentiate for ax dx → 0 dx dx → 0 dx dx → 0 dx
n
as usual. dy dv du
and = u + v
dx dx dx
(b) y = √1 – 5x d dv du
1 or dx (uv) = u dx + v dx
2
y = (1 – 5x)
dy = (1 – 5x) 1 2 – 1 (–5) 3. The first derivative of the function y = u such
1
v
dx 2 that u = f(x) and v = g(x) is obtained by using the
5 – 1
= – (1 – 5x) 2 du dv
2 d u v dx – u dx
1 2
= – 5 formula dx v = v 2 .
2√1 – 5x
247
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