In Boas mathematical methods it says that $$\nabla \cdot [ e_{r}F(r)] $$ is a function of only r for any F(r). I've tried to investigate why this is but I can't seem to get it right. When I do $$\nabla \cdot [ e_{r}F(r)] = F(r)(\nabla\cdot e_{r}) + e_{r}~\cdot \nabla F(r)$$
$$e_{r}=(cos(\theta)sin(\phi),~sin(\theta)sin(\phi),~0)$$
and
$$\nabla \cdot e_r = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 (cos(\theta)sin(\phi) )\right)= \frac{2cos(\theta)sin(\phi)}{r}$$ $$e_{r}~\cdot \nabla F(r) = (cos(\theta)sin(\phi)\frac{\partial F}{\partial r})$$ and I'm not sure how these combine to make this only a function of r... any help would be greatly appreciated
$\require{cancel}$ Define the field ${\bf A} = {\bf e}_r F(r)$, in spherical coordinates
\begin{eqnarray} \nabla\cdot{\bf A} &=& {1 \over r^2}{\partial \left( r^2 A_r \right) \over \partial r} + {1 \over r\sin\theta}{\partial \over \partial \theta} \left( \color{blue}{\cancelto{0}{A_\theta}}\sin\theta \right) + {1 \over r\sin\theta}{\partial \color{blue}{\cancelto{0}{A_\varphi}} \over \partial \varphi} \\ &=& \frac{1}{r^2}\frac{\partial}{\partial r}(r^2 F(r)) \end{eqnarray}
which is a function of just $r$