Multivariable Chain Rule

Question

Does anyone know how to use the multivariable chain rule to solve the following problem?

If $G(x^2+y^2, x+y)=(7x+3y, x+5\,y)$, knowing that $G:\mathbb{R}^2\rightarrow \mathbb{R}^2$, find $(G^{-1})'(24,8)$

Answer 1

Let $f:\mathbb{R}^2\longrightarrow\mathbb{R}^2$ defined by $(x,y)\mapsto(7x+3y,x+5y)$, observe $$G^{-1}\circ f(x,y)=(x^2+y^2,x+y)$$ applying Chain Rule it follows \begin{align} DG^{-1}[f(x,y)]Df(x,y)&=\begin{bmatrix}2x&2y\\1&1\end{bmatrix}\\ DG^{-1}[f(x,y)]\begin{bmatrix}7&3\\1&5\end{bmatrix}&=\begin{bmatrix}2x&2y\\1&1\end{bmatrix} \end{align} Then, putting $x=3,y=1$ and since $f(3,1)=(24,8)$ we get \begin{align} DG^{-1}(24,8)\begin{bmatrix}7&3\\1&5\end{bmatrix}&=\begin{bmatrix}6&2\\1&1\end{bmatrix}\\ DG^{-1}(24,8)&=\begin{bmatrix}6&2\\1&1\end{bmatrix}\begin{bmatrix}7&3\\1&5\end{bmatrix}^{-1}\\ &=\begin{bmatrix}6&2\\1&1\end{bmatrix}\begin{bmatrix}\frac{5}{32}&-\frac{3}{32}\\-\frac{1}{32}&\frac{7}{32}\end{bmatrix}\\ &=\begin{bmatrix}\frac{7}{8}&-\frac{1}{8}\\\frac{1}{8}&\frac{1}{8}\end{bmatrix} \end{align}