Differentiation Short Questions (Question 11 – 14)


Question 11:
Given that the graph of function f ( x ) = h x 3 + k x 2  has a gradient function f ( x ) = 12 x 2 258 x 3 such that h and kare constants. Find the values of and k.

Solution:
f ( x ) = h x 3 + k x 2 = h x 3 + k x 2 f ( x ) = 3 h x 2 2 k x 3 f ( x ) = 3 h x 2 2 k x 3 But it is given that  f ( x ) = 12 x 2 258 x 3 Hence, by comparison,  3 h = 12          and          2 k = 258 h = 4                                k = 129


Question 12:
Given that  y = x 2 x + 3 , show that  d y d x = x 2 + 6 x ( x + 3 ) 2 Find  d 2 y d x 2  in the simplest form .
Solution:
y = x 2 x + 3 d y d x = ( x + 3 ) ( 2 x ) x 2 .1 ( x + 3 ) 2 = 2 x 2 + 6 x x 2 ( x + 3 ) 2 d y d x = x 2 + 6 x ( x + 3 ) 2  (shown) d 2 y d x 2 = ( x + 3 ) 2 ( 2 x + 6 ) ( x 2 + 6 x ) .2 ( x + 3 ) ( x + 3 ) 4 d 2 y d x 2 = ( x + 3 ) [ ( x + 3 ) ( 2 x + 6 ) 2 ( x 2 + 6 x ) ] ( x + 3 ) 4 d 2 y d x 2 = [ 2 x 2 + 6 x + 6 x + 18 2 x 2 12 x ] ( x + 3 ) 3 d 2 y d x 2 = 18 ( x + 3 ) 3


Question 13:
If  y = x 2 + 4 x , show that  x 2 d 2 y d x 2 2 x d y d x + 2 y = 0.
Solution:
y = x 2 + 4 x d y d x = 2 x + 4 d 2 y d x 2 = 2 x 2 d 2 y d x 2 2 x d y d x + 2 y = x 2 ( 2 ) 2 x ( 2 x + 4 ) + 2 ( x 2 + 4 x ) = 2 x 2 4 x 2 8 x + 2 x 2 + 8 x = 0  (Shown)


Question 14:
Given y = x (6 – x), express y d 2 y d x 2 + x d y d x + 18 in terms of xin the simplest form.

Hence, find the value of xwhich satisfies the equation  y d 2 y d x 2 + x d y d x + 18 = 0

Solution:
y = x ( 6 x ) = 6 x x 2 d y d x = 6 2 x d 2 y d x 2 = 2 y d 2 y d x 2 + x d y d x + 18 = ( 6 x x 2 ) ( 2 ) + x ( 6 2 x ) + 18                                     = 12 x + 2 x 2 + 6 x 2 x 2 + 18                                     = 6 x + 18 y d 2 y d x 2 + x d y d x + 18 = 0                6 x + 18 = 0                               x = 3

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