9.7 Pembezaan, SPM Praktis (Kertas 1)


9.7 Pembezaan, SPM Praktis (Kertas 1)

Soalan
 1:
Bezakan ungkapan 2x (4x2 + 2x – 5) terhadap x.

Penyelesaian:
2x (4x2 + 2x – 5) = 8x3 + 4x2– 10x
d/dx (8x3 + 4x2 – 10x)
= 24x + 8x –10 


Soalan 2:
Diberi y= x 3 +2 x 2 +1 3x , cari  dy dx .

Penyelesaian:
y= x 3 +2 x 2 +1 3x y= x 3 3x + 2 x 2 3x + 1 3x y= x 2 3 + 2x 3 + 1 3 x 1 dy dx = 2x 3 + 2 3 1 3 x 2 dy dx = 2x 3 + 2 3 1 3 x 2



Soalan 3:
Diberi y= 3 5x+1 , cari  dy dx

Penyelesaian:
y = 3 5 x + 1 = 3 ( 5 x + 1 ) 1 2 d y d x = 1 2 .3 ( 5 x + 1 ) 3 2 ( 5 ) d y d x = 15 2 [ ( 5 x + 1 ) 3 ] 1 2 d y d x = 15 2 ( 5 x + 1 ) 3


Soalan 4: 
Cari  ds dt  bagi setiap fungsi yang berikut. (a) s= ( t 3 t ) 2 (b) s= ( t+1 )( 35t ) t 2
 

Penyelesaian:
(a)
s = ( t 3 t ) 2 s = ( t 3 t ) ( t 3 t ) s = t 2 6 + 9 t 2 s = t 2 6 + 9 t 2 d s d t = 2 t 18 t 3 = 2 t 18 t 3 d s d t = 2 t 18 t 3

(b)
s = ( t + 1 ) ( 3 5 t ) t 2 s = 3 t 5 t 2 + 3 5 t t 2 s = 5 t 2 2 t + 3 t 2 s = 5 2 t + 3 t 2 s = 5 2 t 1 + 3 t 2 d s d t = 2 t 2 6 t 3 d s d t = 2 t 2 6 t 3



Soalan 5:
Diberi y= 3 5 u 5 , dengan keadaan u=4x+1.  Cari   dy dx  dalam sebutan x.  

Penyelesaian:
y = 3 5 u 5 , u = 4 x + 1 y = 3 5 ( 4 x + 1 ) 5 d y d x = 5. 3 5 ( 4 x + 1 ) 4 .4 d y d x = 12 ( 4 x + 1 ) 4


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