Page 37 - Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation
–3
(b) y = 5 f'(x) = –12x – 8
4x 2 = – 12 – 8
5
y = x x 3
–2
4
dy = –2 3 5 x –2 – 1 4 REMEMBER!
dx 4 Terms are simplified to the form of ax before differentiation
n
5
= – x is done.
–3
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2
= – 5
2x 3 Try Questions 1 – 4 in ‘Try This! 2.2’
(c) y = 3√x
1
y = 3x 2 C Determining the first derivative of
dy = [3x 1 2 – 1 ] composite function
1
dx 2 1 1. A composite function is a function with brackets
3
–
2
= x that is impossible or not easy to expand.
2
= 3 2. The first derivative of composite function
2√x y = u such that u = f(x) and n = integer is done
n
by using chain rule, that is
5 dy dy du
Differentiate each of the following function with dx = du × dx
respect to x.
(a) f(x) = 4x – 8x + 3 3. The limit idea is used to prove the truth of chain
3
(b) f(x) = (5x – 2) rule such as:
2
2
(c) f(x) = 6x – 8x 4 dy lim dy
x 3 dx = dx → 0 dx
Solution lim dy du
(a) f(x) = 4x – 8x + 3 = dx → 01 du × dx 2
3
f'(x) = 3(4x 3 – 1 ) – 1(8x 1 – 1 ) + 0(3x 0 – 1 ) lim dy lim du
= 12x – 8x + 0 = dx → 0 du × dx → 0 dx
2
0
= 12x – 8
2
dy = lim dy × lim du
dx du → 0 du dx → 0 dx
(b) f(x) = (5x – 2) dy dy du
2
2
= (5x – 2)(5x – 2) dx = du × dx
2
2
= 25x – 10x – 10x + 4
4
2
2
= 25x – 20x + 4 3. Since dy = nu n – 1 , then dy = nu n – 1 du .
2
4
1 2
f'(x) = 100x – 40x du dx dx
3
1 2
Form 5
The equation dy = nu n – 1 du is seen as the
REMEMBER! dx dx
formula taken from the chain rule which is often
• The derivative of a constant = 0.
• Terms with brackets which can be expanded should used for composite function.
be expanded first before differentiation is done.
6
(c) f(x) = 6x – 8x 4 Differentiate each of the following function with
x 3 respect to x by using the chain rule.
= 6x – 8x 4 4
x 3 x 3 (a) y = (3x + 2)
6
= 6 – 8x (b) y = 5(1 – 3x)
x 2
= 6x – 8x
–2
246