Prove that for the radially symmetric approach $u(x) = v(r)$ the following is valid Let $x \in \mathbb{R}^n, |x|=r$. Then
$$\Delta u(x)=v''(r)+\frac{n-1}{r}v'(r)$$
So Laplacian of $u$ is equal to $\text{div } \text{grad } u $, that is
$$\text{div } \left(\frac{\partial}{\partial x_1}u(x),\dots,\frac{\partial}{\partial x_n} u(x) \right) = \left(\frac{\partial^2}{\partial x_1^2}u(x)+ \dots +\frac{\partial^2}{\partial x_1 \partial x_n} u(x), \dots \right)$$ So I see that for every row in that vector there's one term where there is a double differentiation of a variable and $(n-1)$ terms with mixed partial differentiations.
Now
$$\frac{\partial}{\partial x}v(|x|) = \frac{x}{|x|}v'(|x|)$$
and
$$\frac{\partial^2}{\partial x^2}v(|x|) = v''(|x|)$$
I see that there is a pattern and that the $|x|$ cancel each other out in the equation I need to prove but I'm stuck on finishing my proof.
 A: I assume you are writing $u(x) = v(\vert x \vert)$. The identity is obtained by a direct computation. Since $$\frac{\partial}{\partial x_i} \vert x \vert = \frac{x_i} {\vert x \vert},$$ it follows from the chain rule that $$ \frac{\partial u}{\partial x_i}=v'(\vert x \vert )\frac{x_i} {\vert x \vert} $$ (Note you have written $\frac{\partial u}{\partial x}=v'(\vert x \vert )\frac{x} {\vert x \vert}$ which doesn't make sense since $x$ is a vector. Maybe you are using that notation $\frac{\partial u}{\partial x}:=\nabla u$?) Next, use the product rule and the chain rule to obtain \begin{align*}
\frac{\partial^2 u}{\partial x_i^2} &=\frac{x_i} {\vert x \vert}\frac{\partial }{\partial x_i} v'(\vert x \vert )+ \frac{v'(\vert x \vert)}{\vert x \vert} \frac{\partial }{\partial x_i}x_i+x_i v'(\vert x \vert) \frac{\partial }{\partial x_i} \frac1 {\vert x \vert} \\
&= \frac{x_i^2}{\vert x \vert^2} v''(\vert x \vert)+\frac{v'(\vert x \vert)}{\vert x \vert} - \frac {x_i^2}{\vert x \vert^3}v'(\vert x\vert). 
\end{align*} Finally, \begin{align*}\Delta u &= \sum_{i=1}^n\frac{\partial^2 u}{\partial x_i^2}  \\
&=\sum_{i=1}^n \bigg [ \frac{x_i^2}{\vert x \vert^2} v''(\vert x \vert)+\frac{v'(\vert x \vert)}{\vert x \vert} - \frac {x_i^2}{\vert x \vert^3}v'(\vert x\vert)\bigg ] \\
&= v''(\vert x \vert)+ \frac{n-1}{\vert x \vert } v'(\vert x \vert)
\end{align*} as required.

Edit: To answer your question as to why $$ \sum_{i=1}^n \bigg [ \frac{v'(\vert x \vert)}{\vert x \vert} - \frac {x_i^2}{\vert x \vert^3}v'(\vert x\vert)\bigg ] 
= \frac{n-1}{\vert x \vert } v'(\vert x \vert).$$ Since $v'$ and $\vert x\vert$ do not depend on $i$, \begin{align*}
\sum_{i=1}^n \bigg [ \frac{v'(\vert x \vert)}{\vert x \vert} - \frac {x_i^2}{\vert x \vert^3}v'(\vert x\vert)\bigg ] &= \sum_{i=1}^n  \frac{v'(\vert x \vert)}{\vert x \vert} - \sum_{i=1}^n\frac {x_i^2}{\vert x \vert^3}v'(\vert x\vert)\\
&=\frac{v'(\vert x \vert)}{\vert x \vert} \sum_{i=1}^n 1  - \frac 1{\vert x \vert^3}v'(\vert x\vert)\sum_{i=1}^nx_i^2. 
\end{align*} Using that $\sum_{i=1}^n 1 = n$ and $\sum_{i=1}^n x_i^2=\vert x \vert^2$, we obtain \begin{align*}
\sum_{i=1}^n \bigg [ \frac{v'(\vert x \vert)}{\vert x \vert} - \frac {x_i^2}{\vert x \vert^3}v'(\vert x\vert)\bigg ] &=n\frac{v'(\vert x \vert)}{\vert x \vert}   - \frac 1{\vert x \vert}v'(\vert x\vert) =\frac {(n-1)}{\vert x \vert}v'(\vert x\vert) 
\end{align*}
