Trying to prove the following identity So I am trying to prove that:
$$\nabla \cdot (f \cdot \vec{r}\,) = \frac{(r^{3}f)'}{r^{2}}$$
where $f=f(r)$ and $\vec{r}=r \hat{r}$ and $\hat{r}$ is the unit vector in the radial direction in spherical coordinates, and del is the del operator. 
I have tried to use the definition of the del operator in spherical coordinates and just plug it in the equation, namely:
$$
\Bigl( \frac{\partial}{\partial r} \hat{r} \Bigr) (f(r) \cdot r \hat{r}) 
= \frac{\partial}{\partial r} (f(r)r) = r \frac{d}{dr}f + 1 \cdot f(r).
$$
However this does not match the right hand side of the equation I am trying to prove. 
Thanks
 A: $\newcommand{\vect}[1]{{\bf #1}}$
In spherical coordinates
$$
\nabla \cdot \vect{F} = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2F_r) + \frac{1}{r\sin\theta}\frac{\partial}{\partial \theta}(\sin\theta A_\theta) + \frac{1}{r\sin\theta} \frac{\partial}{\partial \phi}A_\phi
$$
In your case $\vect{F} = f(r)\vect{r} = f(r)r\hat{\vect{r}}$, so that
$$
\nabla \cdot \vect{F} = \frac{1}{r^2}\frac{{\rm d}}{{\rm d}r}(r^2 f(r)r) = \frac{1}{r^2}\frac{{\rm d}}{{\rm d}r}(r^3 f(r)) 
$$
A: 
I thought it would be instructive to present a coordinate-free approach.  To that end, we proceed.

Using the product rule, $\nabla \cdot (\phi \vec A)=\phi (\nabla\cdot \vec A)+(\nabla \phi)\cdot \vec A$, we have with $\vec A=\vec r$ and $\phi=f$
$$\nabla\cdot(f\vec r)=f(\nabla\cdot\vec r)+(\nabla f)\cdot \vec r$$

Then, noting that $\nabla\cdot\vec r=3$ and $(\nabla f)\cdot \vec r=r\frac{\partial f}{\partial r}$ reveals
$$\nabla\cdot(f\vec r)=3f+r\frac{\partial f}{\partial r}=3f+rf'$$

Finally, using the product rule, we see that $\frac{(r^3f)'}{r^2}=3f+rf'$ thereby establishing the equality
$$\nabla\cdot(f\vec r)=\frac{(r^3f)'}{r^2}$$
as was to be shown!
