Page 47 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan
13. Tentukan terbitan pertama bagi fungsi berikut.
Determine the first derivative for the following functions. SP 2.2.4 TP4

(1 + x) 3
y =
(3 – 2x) 2 (1 + x) 3
Tulis y = (3 – 2x) 2
Katakan u = (1 + x) v = (3 – 2x) 2 Write = (1 + x) (3 – 2x) –2
3
3
Let du dv
= 3(1 + x) , = –4(3 – 2x) Guna/Use y = uv
2
dx dx
dy (3 – 2x)(1 + x) [3(3 – 2x) + 4(1 + x)] dy = u dv + v du
2
= dx dx dx
dx (3 – 2x) 4 dy = (1 + x) (–2)(3 – 2x) (–2) + (3 – 2x) (3)(1 + x) 2
3
–3
–2
(1 + x) (9 – 6x + 4 + 4x) dx
2
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= (1 + x) (4) 3(1 + x) 2
3
(3 – 2x) 3 = (3 – 2x) 3 + (3 – 2x) 2
(1 + x) [13 – 2x] 4(1 + x) + 3(1 + x) (3 – 2x)
2
3
2
= =
(3 – 2x) 3 (3 – 2x) 3
(1 + x) [4(1 + x) + 3(3 – 2x)]
2
=
(3 – 2x) 3
2
Tip Penting (1 + x) (13 – 2x)
= 3
u (3 – 2x)
Guna/Use y =
v
du dv
v – u
dy dx dx
=
dx v 2
–x 3 4 – 3x 2
(a) y = (b) y =
(1 + 2x) 2 2x + x 3
3
2
2
Katakan/Let u = –x , v = (1 + 2x) 2 dy = (2x + x )(–6x) – (4 – 3x )(2 + 3x )
3
du = –3x , dv = 4(1 + 2x) dx (2x + x )
3 2
2
dx dx –12x – 6x – [8 + 12x – 6x – 9x ]
2
4
2
2
4
2
3
dy (1 + 2x) (–3x ) + x (4)(1 + 2x) = (2x + x )
2
3 2
=
dx (1 + 2x) 4 –12x – 6x – 8 – 12x + 6x + 9x 4
2
2
4
2
2
x (1 + 2x)[–3(1 + 2x) + 4x] = 3 2
= (2x + x )
(1 + 2x) 4 –18x + 3x – 8
4
2
x [–2x – 3] = 3 2
2
= (2x + x )
(1 + 2x) 3
(x – 2) 3 
x – 1

(c) y = (d) y =
(1 + x) 5x + 1
Katakan/Let u = (x – 2) , v = 1 + x Katakan/Let u = 
x – 1, v = 5x + 1
3
du = 3(x – 2) , dv = 1 du 1 dv
,
2
dx dx dx = 2 dx = 5
x – 1
dy (1 + x)(3)(x – 2) – (x – 2) 3
2
1
= (5x + 1)   1  – 5
x – 1
dx (x + 1) 2 dy 2 
x – 1
(x – 2) [3(1 + x) – x + 2] dx = (5x + 1) 2
2
=
(1 + 2) 2 5x+ 1 – 10(x – 1)

(x – 2) [5 + 2x] = 2 2
2
x – 1 (5x + 1)
=
(1 + x) 2 11 – 5x
=
2 2
x – 1 (5x + 1)
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02 Hybrid PBD Mate Tamb Tg5.indd 26 09/11/2021 9:24 AM