Page 29 - Ranger SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation

Example 5 Solution
Find Given u = 3x + 1, then du = 3
2
(a) the first derivative for 3x – 2 with dy dx
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4
4
respect to x y = 5u , then du = 5(4)u 3
d 2 = 20u 3
x
(b)   + 
dx x By using the chain rule,
1
(c) yʹ(x) if y(x) = x – 2 dy dy du
4
2 3x 3 dx = du × dx
3
Solution = 20u × 3
= 60u 3 Substitute the expression of u.
d 3x – 2 d 3x 2 1
2
(a)   =  –  = 60(3x + 1) 3
dx 4 dx 4 2
= 3 (2)x 2 – 1 – 0
4 Example 7
= 3x
2 Differentiate each of the following with
d 2 d 1 respect to x.
2
x
(b)   +  = x + 2x  (a) y = (2x – 3x) 3
–1
2
dx x dx (b) y = (3 – x) (x + 2) 3
2
= 1 x 1 2 – 1 + 2(–1)x –1 – 1
2 Solution
1
–
3
2
= 1 x – 2x –2 (a) Let u = 2x – 3x and y = u
2
2 du dy
= 1 – 2 Then dx = 4x – 3 and du = 3u 2
2x x 2 dy dy du
= ×
d 1 2 dx du dx
2
(c) yʹ(x) =  x – x –3  = 3u (4x – 3)
4
dx 2 3 = 3(2x – 3x) (4x – 3)
2
2
1 2
= (4)x 4 – 1 – (–3)x –3 – 1
2 3 (b) Let u = (3 – x) and v = (x + 2) 3
2
= 2x + 2 Then du = 2(3 – x)(–1)
3
x 4 dx
= –2(3 – x)
dv = 3(x + 2) (1)
2
dx
= 3(x + 2)
2
TIPS By using the multiplication rule,
Form 5 Before differentiate, convert the expression in dy Factorize the
the form
dx
1
1
common terms
dv
to x
p
p
(ii) x to x
(i)
–p
x
= u
p
dx + v du (3 – x)(x +
dx
2) from this
2
= (3 – x) (3)(x + 2) expression.
2
2
Example 6 + (x + 2) (–2)(3 – x)
3
2
dy = (3 – x)(x + 2) [3(3 – x) – 2(x + 2)]
Given u = 3x + 1 and y = 5u , find dx = (3 – x)(x + 2) (5 – 5x)
4
2
in terms of x. = 5(3 – x)(x + 2) (1 – x)
2
182
02 Ranger Mate Tambahan Tg5.indd 182 25/02/2022 9:23 AM