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

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 $u=1+x^2+y^2$, $v=1+x^2{y}^2$.

Apply the product rule and the chain rule once more to each expression in part (i). The second derivative chain rule givesandFor example, the coefficient of $f_{uv}$ in the mixed partial derivative is $u_x{v}_y+v_x{u}_y=4x^{3}y+4x{y}^3$, while the final term is $v_{xy}{f}_v=4xy{f}_v$. The symmetry $f_{uv}=f_{vu}$ is used in combining the mixed terms under the usual twice-continuously-differentiable interpretation.

For $f(u,v)=\ln u+\ln v$, the needed partial derivatives areSubstituting these into part (ii) givesSince $v-x^2{y}^2=1$, the last two terms combine. Thus

First express the function in the original variables:The chain rule givesTaking its partial derivative with respect to $y$ yieldsThe numerator in the second fraction reduces to $4xy$, reproducingBoth logarithms are defined for every real $x,y$ because their arguments are at least one.