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 k are constants. Find the values of h and k.

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


Question 12:
Given that y= x 2 x+3 , show that  dy dx = x 2 +6x ( x+3 ) 2 Find  d 2 y d x 2  in the simplest form.
Solution:
y= x 2 x+3 dy dx = ( x+3 )( 2x ) x 2 .1 ( x+3 ) 2 = 2 x 2 +6x x 2 ( x+3 ) 2 dy dx = x 2 +6x ( x+3 ) 2  (shown) d 2 y d x 2 = ( x+3 ) 2 ( 2x+6 )( x 2 +6x ).2( x+3 ) ( x+3 ) 4 d 2 y d x 2 = ( x+3 )[ ( x+3 )( 2x+6 )2( x 2 +6x ) ] ( x+3 ) 4 d 2 y d x 2 = [ 2 x 2 +6x+6x+182 x 2 12x ] ( x+3 ) 3 d 2 y d x 2 = 18 ( x+3 ) 3

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


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

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

Solution:
y=x( 6x )=6x x 2 dy dx =62x d 2 y d x 2 =2 y d 2 y d x 2 +x dy dx +18=( 6x x 2 )( 2 )+x( 62x )+18                                    =12x+2 x 2 +6x2 x 2 +18                                    =6x+18 y d 2 y d x 2 +x dy dx +18=0               6x+18=0                              x=3