Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 1 / Solution

Apply the chain rule twice: once to the square and once to the argument $x^3$ of the sine. The inner derivative is $y^{\prime }=3x^2\cos (x^3)-2\sin (2x)$. ThereforeThe original PDF has $\sin (x^3)$, rather than the cube of $\sin x$.