Answer
$$y'=2x\sin^2(2x^2)+4x^3\sin(4x^2)$$
Work Step by Step
$$y=x^2\sin^2(2x^2)$$
Applying the Product Rule, we have $$y'=(x^2)'\sin^2(2x^2)+x^2(\sin^2(2x^2))'$$
Now, apply the Chain Rule for $(\sin^2(2x^2))'$: $$y'=2x\sin^2(2x^2)+x^2(2\sin(2x^2))(\sin(2x^2))'$$
Again, we apply the Chain Rule for $(\sin(2x^2))'$: $$y'=2x\sin^2(2x^2)+x^2\Big(2\sin(2x^2)\Big)\Big(\cos(2x^2)\Big)(2x^2)'$$ $$y'=2x\sin^2(2x^2)+x^2\Big(2\sin(2x^2)\Big)\Big(\cos(2x^2)\Big)\times4x$$ $$y'=2x\sin^2(2x^2)+8x^3\sin(2x^2)\cos(2x^2)$$
Recall that $2\sin A\cos A=\sin2A$: $$y'=2x\sin^2(2x^2)+4x^3\sin(4x^2)$$
