I have the following expression $$ \textrm{div} \, \frac{\vec{r} - \vec{a}}{\lvert\vec{r} - \vec{a}\rvert^n},\quad n \in \mathbb{N} $$ where $\vec{r} \neq \vec{a}$ and $\vec{a}$ is a constant vector ($\vec{r}$ is not). I need to simplify it using the divergence product rule, namely $$ \textrm{div} \, (f\vec{A}) = f \, \textrm{div}(\vec{A}) + (\textrm{grad} \, f) \cdot \vec{A} $$
My approach so far:
Notice that the expression $\vec{r} - \vec{a}$ represents a vector field, as $\vec{r}$ is variable and the expression returns another vector. Likewise, $\lvert \vec{r} - \vec{a}\rvert^{-n}$ can be seen as a scalar field, taking in a vector and returning a scalar.
Therefore, set $f = \lvert \vec{r} - \vec{a}\rvert^{-n}$ and $\vec{A} = \vec{r} - \vec{a}$, then we have: $$\begin{align*} \textrm{div} \, (f \vec{A}) &= f \, \textrm{div}(\vec{A}) + (\textrm{grad} \, f) \cdot \vec{A} \\ &= \lvert \vec{r} - \vec{a}\rvert^{-n} \; \textrm{div}(\vec{r} - \vec{a}) + (\textrm{grad} \, \lvert \vec{r} - \vec{a}\rvert^{-n}) \cdot (\vec{r} - \vec{a}) \end{align*} $$ Now I tried to simplify the subexpressions first, starting with $$\begin{align*} \textrm{div}(\vec{r} - \vec{a}) &= \sum_{i=1}^n \frac{\delta (r_i - a_i)}{\delta x_i} \\ &= \sum_{i=1}^n \frac{\delta r_i}{x_i} \\ &= \textrm{div}\,{\vec{r}} \end{align*} $$ However, I didn't get far simplyfing the gradient term. Is my approach wrong or is there another way to simplify this expression?