Can anyone show me how to get the first and second derivative of y=sin(t^2)? I need a refreshing on using the chain rule. Solution dy/dt=d/dt(sint^2) =cos(t^2)d/dt(t^2) =cos(t^2)(2t) =2tcos(t^2) d(uv)=udv+vdu d^2y/dt^2= 2td/dt(cos(t^2))+2cos(t^2)d/dt(t) =2t(-sin(t^2)d/dt(t^2)+2cos(t^2)*1 =- 2tsin(t^2)2t+2cos(t^2) =-4t^2sin(t^2)+2cos(t^2).