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2.10.1 Differentiation Short Questions (Question 1 – 5)


Question 1:
Differentiate the expression 2x (4x2 + 2x – 5) with respect to x.

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


Question 2:
Given that  y = x 3 + 2 x 2 + 1 3 x ,  find  d y d x . .

Solution:
y = x 3 + 2 x 2 + 1 3 x y = x 3 3 x + 2 x 2 3 x + 1 3 x y = x 2 3 + 2 x 3 + 1 3 x − 1 d y d x = 2 x 3 + 2 3 − 1 3 x − 2 d y d x = 2 x 3 + 2 3 − 1 3 x 2

Question 3:
Given that  y = 3 5 x + 1 ,  find  d y d x

Solution:
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

Question 4:
Find d s d t for each of the following functions.
( a )   s = ( t − 3 t ) 2 ( b )   s = ( t + 1 ) ( 3 − 5 t ) t 2

Solution:
(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

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


Question 5:
Given that  y = 3 5 u 5 , where u = 4x + 1. Find d y d x in terms of x.

Solution:
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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