Jacobian of $f(|x|)x$ Suppose one has a $C^1$ function $f:\mathbb{R} \rightarrow [0,\infty)$. A throwaway line in a paper I'm reading claims 
\begin{equation}
\det \left[ \nabla \left( f(|\vec{x}|)\vec{x} \right) \right] = f^{n-1}(|\vec{x}|) \left[f(|\vec{x}|) + f'(|\vec{x}|)|\vec{x}| \right]
\end{equation}
for $\vec{x} \in \mathbb{R}^n$. Any ideas on how to prove this?

One could brute-force this by computing the normal and tangential derivatives of $f$. But I prefer the following argument. 


*

*$f$ is radially symmetric; that is, $f\circ T=T\circ f$ for all orthogonal transformations $T$. 

*Therefore, its Jacobian determinant depends only on $|x|$. 

*A function that depends only on $|x|$ is determined by its integrals over balls $B_r=\{x:|x|<r\}$

*The integral of Jacobian determinant over $B_r$ is the volume of the sphere bounded by $f(\partial B_r)$. (This is easy to see when $f$ is increasing, and follows from the change of variables formula in general). By definition $f(\partial B_r)$ has radius $rf(r)$.

*Thus, the integral of Jacobian over the sphere $|x|=r$ is 
$$\frac{d}{dr}(n^{-1}\omega_{n-1} (rf(r))^{n}) 
=\omega_{n-1}f(r)^{n-1} r^{n-1} ( f(r) +rf'(r))$$

*Divide by $\omega_{n-1} r^{n-1}$ (which is the area of the sphere $|x|=r$) to obtain the claimed result.

