A simple question about the laplacian

Question

Suppose that $f:X\subseteq R^n\to R$ depends only on the distance $(x_1,x_2,...,x_n)$ is from the origin in $R^n$ (i.e. $f(\vec x)=g(r)$ where $r=\left | \vec x \right |$) Show that for all $\vec x\ne 0$ the laplacian is given by:$$\nabla^2f=\frac{n-1}{r}g'(r)+g''(r)$$ I get the solution without $-1$. Am I right? Or have I made a mistake?

Answer 1

If we set $r=r(x)=|x|$,

$\frac{\partial f(|x|)}{\partial x_i}=f'(|x|)\frac{x_i}{|x|}$

and differentiating another time with respect to $x_i$:

$\frac{\partial^2f(|x|)}{\partial x_i^2}=f''(|x|)\frac{x_i^2}{|x|^2}+\frac{1}{|x|} f'(|x|)-\frac{x_i^2}{|x|^3}f'(|x|)$.

Then summing over $i=1,\ldots,n$ you get

$\nabla^2f=\frac{n-1}{r}f'(r)+g''(r).$