Find the derivative matrix $g\circ f$ at $\dbinom uv$ Vector function $f$ and $g$ are defined by:
$$f \dbinom uv=\begin{pmatrix}u\cos(v)\\u\sin(v)\end{pmatrix},\begin{pmatrix}0<u<\infty\\-\frac{\pi}{2}<v<\frac{\pi}{2}\end{pmatrix}$$
$$g \dbinom xy=\begin{pmatrix}\sqrt{x^2+y^2}\\\arctan(\frac{y}{x})\end{pmatrix}, 0<x<\infty$$
Find the derivative matrix of $g\circ f$ at $\dbinom uv$.
My workings:
$$f' \dbinom uv=\begin{pmatrix}\cos(v) & -u\sin(v)\\\sin(v)&u\cos(v)\end{pmatrix}$$
$$g' \dbinom xy=\begin{pmatrix}\frac{x}{\sqrt{x^2+y^2}}& \frac{y}{\sqrt{x^2+y^2}}\\-\frac{y}{x^2+y^2}&\frac{x}{x^2+y^2}\end{pmatrix}$$
I had thought I would have to multiply $f'$ with $g'$ and replace $x,y$ with $u,v$, however I do not get the final expected result which is:
$$\begin{pmatrix}1&0\\0&1\end{pmatrix}$$

As noticed in the comments by chain rule we have
$$(g \circ f)' \dbinom uv = g' \bigg( f \dbinom uv \bigg) f' \dbinom uv = g' \dbinom{u \cos v}{u \sin v} f' \dbinom uv$$
that is in this case
$$\begin{pmatrix}\frac{x}{\sqrt{x^2+y^2}}& \frac{y}{\sqrt{x^2+y^2}}\\-\frac{y}{x^2+y^2}&\frac{x}{x^2+y^2}\end{pmatrix}_{(u\cos v, u \sin v)}\cdot \begin{pmatrix}\cos v & -u\sin v\\\sin v&u\cos v\end{pmatrix}$$
that is
$$\begin{pmatrix}\cos v& \sin v\\-\frac{\sin v}u&\frac{cos v}u\end{pmatrix}\cdot \begin{pmatrix}\cos v & -u\sin v\\\sin v&u\cos v\end{pmatrix}=\begin{pmatrix}1 & 0\\ 0&1\end{pmatrix}$$
Refer also to the related

*

*Derivation of the multivariate chain rule
