How can I compute the laplacian $\nabla^2 g $ where $ g $($\textbf{r}$) = f(r), where f is a function of a single variable, r = |$\textbf{r}$|.
I understand that the laplacian is defined to be $ \frac{\partial^2g}{\partial x_i ^2} $ but I don't really understand the definition of $ \phi $ and how I should apply the chain rule here.
The Laplacian Operator can be written
$$\nabla ^2g(\vec r)=\nabla \cdot \nabla g(\vec r)$$
Now, if $g(\vec r)=f(|\vec r|)=f(r)$, then from the chain rule we have
$$\nabla f(r)=f'(r)\nabla r \tag 1$$
Applying the product rule to $(1)$ yields
$$\nabla^2 f(r)=\nabla f'(r)\cdot \nabla r+f'(r)\nabla^2 r$$
whereupon from the chain rule we have
$$\nabla^2 f(r)=f''(r)\nabla r\cdot \nabla r+f'(r)\nabla^2 r$$
Using $\nabla r=\hat r$ and $\nabla^2 r=2/r$ we find that
$$\bbox[5px,border:2px solid #C0A000]{\nabla^2 f(r)=f''(r)+\frac2r f'(r)}$$
NOTE:
One could use the Laplacian in spherical coordinates to obtain the result more efficiently. There, we write
$$\begin{align} \nabla^2 f(r)&=\frac1{r^2}\frac{\partial }{\partial r}\left(\frac{r^2\partial f(r)}{\partial r}\right)\\\\ &=\frac1{r^2}\frac{\partial }{\partial r}\left(r^2f'\right)\\\\ &=\frac1{r^2}\left(r^2f''+2rf'\right)\\\\ &=f''+\frac{2}{r}f' \end{align}$$
as expected!
