Prove that $\triangledown^2 f(r) = \frac{d^2f}{dr^2}+ \frac{2}{r} \frac{df}{dr} $ $\triangledown^2 f(r) = \frac{d^2f}{dr^2}+ \frac{2}{r} \frac{df}{dr} $ where $r^2 = x^2 + y^2 +z^2$
Now i  write as $\triangledown f(r) = f'(r) \frac{r}{|r|}$ . taking again $\triangledown f'(r) \frac{r}{|r|}$. How do i proceed as r in numerator is vector
 A: Conventions used: $\vec{r} = $ vector and $r =$ magnitude of $\vec{r}$
$$y=\nabla^2f(r) = \nabla\cdot\nabla(f(r)) = \nabla\cdot[f'(r)\nabla(r)] = \nabla\cdot[\color{red}{f'(r)\frac{\vec{r}}{r}}]$$
$$y= f'(r)\nabla\cdot\bigg[\frac{\vec{r}}{r}\bigg]+\frac{\vec{r}}{r}\cdot\nabla f'(r)$$
$$y = f'(r)\frac{r\nabla\cdot{\vec{r} \ - \ \vec{r}\cdot\nabla{r}}}{r^2}+\frac{\vec{r}}{r}\cdot\bigg( \color{red}{f''(r)\frac{\vec{r}}{r}}\bigg)$$
Same as the first step.
$$y = f'(r)\frac{r \cdot3 - \vec{r}\cdot\frac{\vec{r}}{r}}{r^2}+\frac{r^2}{r^2}f''(r)$$
$$y = f'(r)2\frac{r}{r^2}+ f''(r)$$

$$y = f''(r)+\frac{2}{r}f'(r)$$

A: I'm going to write $x$ for an element of $\mathbb{R}^n$ and $r=\|x\|,$ if that's okay.
You correctly found that $\nabla f(r)=f'(r)\frac{x}{r}.$ By the product rule, the expression you want for $\nabla^2 f(r)$ is $$\nabla^2 f(r)=\nabla (f'(r))\cdot \frac{x}{r}+f'(r) \nabla\cdot \left(\frac{x}{r}\right).$$ Now, just compute these (spoiler below).

 I'll do this in $\mathbb{R}^n$, then you can immediately specialize to $\mathbb{R}^3)$. Compute $$\nabla f'(r)\cdot\frac{x}{r}=f''(r)\frac{x}{r}\cdot \frac{x}{r}=f''(r),$$ and $$f'(r) \nabla\cdot \left(\frac{x}{r}\right)=f'(r) \frac{n-1}{r}.$$ If you take $n=3$ (i.e. if you're working in $\mathbb{R}^3$, which you are), then you have your result.

