Question 976800
{{{dy/dx=(dy/dt)/(dx/dt)}}}
{{{dy/dt=(2at)}}}'={{{2a}}}
{{{dx/dt=(at^2)}}}'={{{2at}}}
{{{dy/dx=(2a)/(2at)=1/t}}}
{{{d(dy/dx)/dt=(1/t)}}}'={{{-1/t^2}}}
{{{d^2y/dx^2 =(d(dy/dx)/dt)/(dx/dt)}}}
{{{d^2y/dx^2 =(-1/t^2)/(2at)=-1/(2at^3)}}}
Answer: b)