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Bab 9 Pembezaan

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9.7 Pembezaan, SPM Praktis (Kertas 1)
Soalan 6:
Diberi f (x) = 3x2(4x2 – 1)7, cari f’(x).

Penyelesaian:
f (x) = 3x2(4x2 – 1)7
f’(x) = 3x2. 7(4x2 – 1)6. 8x + (4x2 – 1)7. 6x
f’(x) = 168x3 (4x2 – 1)6 + 6x (4x2 – 1)7
f’(x) = 6x (4x2 – 1)6[28x2+ (4x2 – 1)]
f’(x) = 6x (4x2 – 1)6 (32x2 – 1)


Soalan 7:
Diberi y = (1 + 4x)3(3x2 – 1)4, cari dy/dx.

Penyelesaian:
y = (1 + 4x)3(3x2 – 1)4
dy/dx
= (1 + 4x)3. 4(3x2 – 1)3.6x + (3x2 – 1)4. 3(1 + 4x)2.4
= 24x (1 + 4x)3(3x2 – 1)3 + 12 (3x2 – 1)4(1 + 4x)2
= 12 (1 + 4x)2(3x2 – 1)3 [2x (1 + 4x) + (3x2 – 1)]
= 12 (1 + 4x)2(3x2 – 1)3 [2x + 8x2 + 3x2 – 1]
= 12 (1 + 4x)2(3x2 – 1)3 [11x2 + 2x  – 1]


Soalan 8:
Diberi f(x)=3x 4 x 2 −1  , cari f'(x). 

Penyelesaian:
f(x)=3x 4 x 2 −1 =3x ( 4 x 2 −1 ) 1 2 f'(x)=3x. 1 2 ( 4 x 2 −1 ) − 1 2 .8x+ ( 4 x 2 −1 ) 1 2 .3 f'(x)=12 x 2 ( 4 x 2 −1 ) − 1 2 +3 ( 4 x 2 −1 ) 1 2 f'(x)=3 ( 4 x 2 −1 ) − 1 2 [ 4 x 2 +( 4 x 2 −1 ) ] f'(x)= 3( 8 x 2 −1 ) ( 4 x 2 −1 )


Soalan 9:
Diberi y= 1−5 x 4 x−3 , cari  dy dx . 

Penyelesaian:
dy dx = v du dx −u dv dx v 2 dy dx = ( x−3 ).−20 x 3 −( 1−5 x 4 ).1 ( x−3 ) 2 dy dx = −20 x 4 +60 x 3 −1+5 x 4 ( x−3 ) 2 dy dx = −15 x 4 +60 x 3 −1 ( x−3 ) 2


Soalan 10:
Diberi f(x)= ( x 2 −3 ) 5 1−3x , cari f'(0). 

Penyelesaian:
f(x)= ( x 2 −3 ) 5 1−3x f'(x)= v du dx −u dv dx v 2 f'(x)= ( 1−3x ).5 ( x 2 −3 ) 4 .2x− ( x 2 −3 ) 5 .−3 ( 1−3x ) 2 f'(x)= 10x( 1−3x ) ( x 2 −3 ) 4 +3 ( x 2 −3 ) 5 ( 1−3x ) 2 f'(x)= ( x 2 −3 ) 4 [ 10x−30 x 2 +3( x 2 −3 ) ] ( 1−3x ) 2 f'(x)= ( x 2 −3 ) 4 [ −27 x 2 +10x−9 ] ( 1−3x ) 2 ∴f'(0)= ( 0 2 −3 ) 4 [ −27 ( 0 ) 2 +10( 0 )−9 ] ( 1−3( 0 ) ) 2 f'(0)= 81×( −9 ) 1 =−729