If y=x ( x 2-3 x ) for 0 < x < 2 3 , then d^2 y d x^2 at x= 1 2 is

Step-by-step solution

Given, y=x ( x 2-3 x ) , for 0 < x < 2 3 On differentiating w.r.t. to ' x ', we are getting aligned d y d x & =x ( 1 x - -3 2-3 x )+ ( x 2-3 x ) d y d x & = 2 2-3 x + ( x 2-3 x ) aligned Again differentiating w.r.t. to ' ', we are getting aligned d^2 y d x^2 & = -2 (2-3 x)^2 (-3)+ ( 1 x - -3 2-3 x ) & = 6 (2-3 x)^2 + 2 x(2-3 x) aligned So, aligned & d^2 y d x^2 ( at x= 1 2 )= 6 (1 / 2)^2 + 2 1 / 2 1 2 & =24+8=32 & aligned