Solution
= Solution
Apply the <chain rule> twice: once to the square and once to the argument $x^3$ of the <sine>. The inner <derivative> is $y'=3x^2\cos(x^3)-2\sin(2x)$. Therefore
$$
\boxed{\frac{d(y^2)}{dx}=2\bigl[\sin(x^3)+\cos(2x)\bigr]\bigl[3x^2\cos(x^3)-2\sin(2x)\bigr].}
$$
The original PDF has $\sin(x^3)$, rather than the cube of $\sin x$.