If $u(r)$ is a scalar function of $r=|\textbf{r}|$, show that:
$\nabla^2u=u''+2u'/r$
I assumed that $\frac{\partial u}{\partial r}$=$\frac{du}{dr}$ , since $u$ is dependent on $r$ alone.
I tried using the chain rule on both sides, however to no avail. I cannot seem to get any consistency with the derivatives.
I'm going to use $r=|{\bf{x}}|,$ if that's okay.
Use the chain rule to find that, generally in $\mathbb{R}^n,$ $$\nabla u=(u')(\nabla r)=u' \frac{{\bf{x}}}{r}$$ and \begin{align*}\nabla^2 u&=\nabla\cdot \left(u' \frac{{\bf{x}}}{r}\right)=(\nabla u)\frac{{\bf{x}}}{r}+u'\nabla\cdot \left(\frac{{\bf{x}}}{r}\right)\\ &=u''\frac{|{\bf{x}}|^2}{r^2}+ \frac{n-1}{r}u'\\ &=u''+\frac{n-1}{r}u'.\end{align*} Taking $n=3$ yields your desired result.
In case you're having trouble taking these gradients, I'll also show how to calculate $\nabla r,$ for example: for $1\leq j\leq n,$ \begin{align*}\partial_j r&=\partial_j\sqrt{x_1^2+x_2^2+\cdots+x_j^2+\cdots +x_2^2}\\ &=\frac{1}{2}\frac{2x_j}{\sqrt{x_1^2+x_2^2+\cdots+x_j^2+\cdots +x_2^2}}\\ &=\frac{x_j}{r}. \end{align*} Since $\nabla r=(u_j)_{j=1}^n,$ this shows that $\nabla r=\frac{{\bf{x}}}{r}.$