Page 46 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan

6
3
3

(b) y =  1 x – x – 1 , x = –1 (c) y =  , x = –3
2
4
3x – 2
2
2
2
3
2
y =  1 x – x – 1  3 y = 6(3x – 2) – 1 2
4
2
2
2
3
 
–
2
2
dy 1 3 2 dy = 6 – 1 (3x – 2) (6x)
2
3
4
= (2x – 3x)(3)  x – x – 1  dx 2
dx 2 2 –18x
dy 1 3 2 = 2 -
(3x – 2)
x = –1, = (–2 + 3)(3)  – – 1   3
dx 2 2 dy –18(–3)
= 3(–2) = 12 x = –3, dx = 125
2
54
=
125
2Reserved.

12. Tentukan terbitan pertama bagi fungsi berikut.
Determine the first derivative for the following functions. SP 2.2.4 TP4
= Penerbitan Pelangi Sdn Bhd. All Rights
(a) y = (x + 1) (1 – x)
–1
y = (x – 1) (2 – x ) Katakan/Let u = (x + 1) , v = (1 – x) 2
–1
2 3
2
du dv
Katakan/Let say u = (x – 1) , v = (2 – x ) . Maka/Hence = –(x +1) , = –2(1 – x)
–2
2
2 3
du dv dx dx
Maka/Hence = 2(x – 1), = –6x(2 – x ) Gunakan/Use
2 2
dx dx
Gunakan/Use dy = u dv + v du
dx
dx
dx
dy = u dv + v du dy (1 – x) (–1)
2
–1
dx dx dx dx = (x + 1) (–2)(1 – x) + (x + 1) 2
dy = (x – 1) (–6x)(2 – x ) + (2 – x ) (2)(x – 1) –2(1 – x) (1 – x) 2
2 3
2
2 2
dx = (x + 1) – (x + 1) 2
= 2(x – 1)(2 – x ) (3x – 4x + 2) 2
2
2 2
= –2(1 – x)(x + 1) – (1 – x)
(x + 1) 2
(1 – x)[–2(x + 1) – (1 – x)]
=
(x + 1) 2
(1 – x)[–x – 3]
=
(x + 1) 2
(x – 1)(x + 3)
=
(x + 1) 2
2 
2
(b) f(x) = x x + 1 (c) y = 2x (x – 3) 3
4
1
x + 1(2x)
4
1
2
f’(x) = x 2   (x + 1) 1 +  Katakan/Let u = 2x , dv v = (x – 3) 2
1
2
du
2
3
2
x + 4x(x + 1) Maka/Hence dx = 8x , dx = 3(2x)(x – 3) 2
2

2 x + 1 Gunakan/Use
2
5x + 4x dy dv du
= = u + v

2 x + 1 dx dx dx
x(5x + 4) dy
= = 2x (3)(2x)(x – 3) + (8x )(x – 3) 3
2
2
3
2
4

2 x + 1 dx
= 4x (x – 3) [3x + 2x – 6]
2
2
2
2
3
= 4x (x – 3) (5x – 6)
2
3
2
2

25 © Penerbitan Pelangi Sdn. Bhd.
02 Hybrid PBD Mate Tamb Tg5.indd 25 09/11/2021 9:24 AM