9.7 Pembezaan, SPM Praktis (Kertas 1)


9.7 Pembezaan, SPM Praktis (Kertas 1)

Soalan 6:
Diberi f (x) = 3x(4x2 – 1)7, cari f’(x).

Penyelesaian:
f (x) = 3x(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)[28x+ (4x2 – 1)]
f’(x) = 6x (4x2 1)6 (32x2 1)


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

Penyelesaian:
y = (1 + 4x)(3x2 – 1)4
dy/dx
= (1 + 4x)3. 4 (3x2 – 1)3. 6x + (3x2 – 1)4. 3(1 + 4x)2.4
= 24x (1 + 4x)(3x2 – 1)3 + 12 (3x2 – 1)4(1 + 4x)2
= 12 (1 + 4x)(3x2 – 1)3 [2x (1 + 4x) + (3x2 – 1)]
= 12 (1 + 4x)(3x2 – 1)3 [2x + 8x2 + 3x2 – 1]
= 12 (1 + 4x)(3x2 – 1)3 [11x2 + 2x  – 1]


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

Penyelesaian:
f ( x ) = 3 x 4 x 2 1 = 3 x ( 4 x 2 1 ) 1 2 f ( x ) = 3 x . 1 2 ( 4 x 2 1 ) 1 2 .8 x + ( 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= 15 x 4 x3 , cari  dy dx .  

Penyelesaian:
d y d x = v d u d x u d v d x v 2 d y d x = ( x 3 ) . 20 x 3 ( 1 5 x 4 ) .1 ( x 3 ) 2 d y d x = 20 x 4 + 60 x 3 1 + 5 x 4 ( x 3 ) 2 d y d x = 15 x 4 + 60 x 3 1 ( x 3 ) 2


Soalan 10:
Diberi f(x)= ( x 2 3 ) 5 13x , cari f'(0).  

Penyelesaian:
f ( x ) = ( x 2 3 ) 5 1 3 x f ( x ) = v d u d x u d v d x v 2 f ( x ) = ( 1 3 x ) .5 ( x 2 3 ) 4 .2 x ( x 2 3 ) 5 . 3 ( 1 3 x ) 2 f ( x ) = 10 x ( 1 3 x ) ( x 2 3 ) 4 + 3 ( x 2 3 ) 5 ( 1 3 x ) 2 f ( x ) = ( x 2 3 ) 4 [ 10 x 30 x 2 + 3 ( x 2 3 ) ] ( 1 3 x ) 2 f ( x ) = ( x 2 3 ) 4 [ 27 x 2 + 10 x 9 ] ( 1 3 x ) 2 f ( 0 ) = ( 0 2 3 ) 4 [ 27 ( 0 ) 2 + 10 ( 0 ) 9 ] ( 1 3 ( 0 ) ) 2 f ( 0 ) = 81 × ( 9 ) 1 = 729


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