Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 14 / b / i / Solution

The chain rule gives $f_x=f_u{u}_x+f_v{v}_x$ and $f_y=f_u{u}_y+f_v{v}_y$. Since $u_x=2x$, $u_y=2y$, $v_x=2x{y}^2$ and $v_y=2x^{2}y$, the answer isEvery partial derivative of $f$ on the right is evaluated at $u=1+x^2+y^2$, $v=1+x^2{y}^2$.