I'm trying to solve the following problem:
Given $\vec{r} = (x,y,z)$ , $r= \lVert \vec{r} \rVert$ and $f:\mathbb{R} \to \mathbb{R}$ a twice differentiable function, show that $$ \Delta f(r) = f''(r) + \frac{2}{r} f'(r) $$
I've already shown previously that $ \nabla f(r) = f'(r) \frac{\vec{r}}{r}$, and I was trying to use the property that $\Delta \varphi = \nabla \cdot \nabla \varphi$ to calculate my problem. From this route and after some use of the chain rule I got to \begin{align*} \Delta f(r) &= \left[ \frac{\partial}{\partial x} \frac{f'(r)}{r} x + \frac{\partial}{\partial y}\frac{f'(r)}{r} y+\frac{\partial}{\partial z} \frac{f'(r)}{r} z\right]=3 \frac{f'(r)}{r} +\frac{\left(\frac{\partial}{\partial x}r + \frac{\partial}{\partial y}r +\frac{\partial}{\partial z}r\right)\left[rf''(r)+f'(r)\right]}{r^2} \\&= 3 \frac{f'(r)}{r}+ \left[ \frac{f''(r)}{r^2} + \frac{f'(r)}{r^3} \right](x+y+z) \end{align*} which I didn't know how to simplify further. I think I may have made a mistake in the computation, but I'm not sure.
The other idea I had was to use the product rule for multiplication by a scalar applied to the divergence, which states that given $\zeta$ a scalar field and $\varphi$ a vector field, then $\nabla \cdot ( \zeta \varphi ) =\zeta \ \nabla \cdot \varphi + ( \nabla \zeta ) \cdot \varphi$. Applying this to the problem results in $$ \Delta f(r) = \frac{f'(r)}{r}\ \nabla \cdot \vec{r} + \left( \nabla\frac{f'(r)}{r} \right) \cdot \vec{r} = 3\frac{f'(r)}{r}\ + \left( \nabla\frac{f'(r)}{r} \right) \cdot \vec{r} $$ but from here it seemed to me that by computing $\left( \nabla\frac{f'(r)}{r} \right)$ I was pretty much repeating the same steps I did in the first method I tried.
Am I on the right track? Or can anyone tell me if there's an easier way to compute this? I would very much appreciate the help. Thank you!
Edit:
I've learned that this problem was incorrectly transcribed, and rather was written for a "2-dimensional gradient" using $\vec{r} =(x,y)$, $r= \lVert \vec{r} \rVert = \sqrt{x^2+y^2}$ and $\Delta \varphi = \frac{\partial^2}{\partial x^2} \varphi+\frac{\partial^2}{\partial y^2} \varphi$. From which we get \begin{align*} &\Delta f(r) = \frac{\partial^2}{\partial x^2} f(r)+\frac{\partial^2}{\partial y^2} f(r) = \frac{\partial}{\partial x} f'(r)\frac{x}{r}+\frac{\partial}{\partial y} f'(r)\frac{y}{r}\\ &=\left[\frac{x}{r}f''(r) \frac{x}{r} + f'(r) \frac{y^2}{r^3}\right] + \left[\frac{y}{r}f''(r) \frac{y}{r} + f'(r) \frac{x^2}{r^3} \right]= f''(r) + \frac{1}{r} f'(r) \end{align*} Nevertheless, I still think the full 3D cartesian expression is worth analyzing.
$r=\sqrt{x^2+y^2+z^2} \implies \frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^2+y^2+z^2}}$ etc. The $x$-component of $$\vec \nabla \cdot\left(\frac{f'(r) \vec r}{r}\right)=\frac{\partial}{\partial x}\left(\frac{f'(r) x}{r}\right)= f''(r)\frac{x^2}{r^2}+ \frac{f'(r)}{r}+f'(r)\frac{-x}{r^2}\frac{x}{r}$$ So adding similar $ x,y,z$ components we get $$\vec \nabla \cdot\left(\frac{f'(r) \vec r}{r}\right)=f''(r)+3\frac{f'(r)}{r}-\frac{f'(r)}{r}=f''(r)+2\frac{f'(r)}{r}$$