derivative $\nabla \frac{\mathbf{r}}{r^k}$ in the context of Geometric Calculus Suppose $\mathbf{r} = \mathbf{x - x'}$ is the position vector in $\mathbf{R^n}$, and $r = |\mathbf{r}| = |\mathbf{x - x'}|$.   Do we have $\nabla \frac{\mathbf{r}}{r^k} = \frac{n-k-1}{r^k}$ or $\nabla \frac{\mathbf{r}}{r^k} = \frac{n-k}{r^k}$? The former is from page 7 in Tutorial on Geometric Calculus  by David Hestenes.  If this is not a typo, it is intriguing that $\nabla \frac{\mathbf{r}}{r^k}$ is not equal to the $\nabla \cdot \frac{\mathbf{r}}{r^k}$ in ordinary vector calculus. 
 A: First of all, if $\mathbf{r}$ is a vector, then on the right you have a divergence, not a gradient (let's find it in a Cartesian coordinates).
$$
\nabla\cdot\frac{\mathbf{r}}{r^k} = \sum\limits_{i=1}^{n}\frac{\mathrm{d}}{\mathrm{d}x_i}\frac{x_i}{\sqrt{x_1^2 + \cdots + x_i^2}^k} = \frac{n-k}{r^k}
$$
Let's find the $\frac{\mathrm{d}}{\mathrm{d}x_i}$ of $\frac{x_i}{r^k}$:
$$
\frac{\mathrm{d}}{\mathrm{d}x_i}\frac{x_i}{\sqrt{x_1^2 + \cdots + x_i^2}^k} = -\frac{k x_i^2}{\sqrt{x_1^2 + \cdots + x_i^2}^{k+2}} + \frac{1}{\sqrt{x_1^2 + \cdots + x_i^2}^k}
$$
Multiplying and simplifying yield the following:
$$
\frac{(1-k)x_i^2 + \sum\limits_{j \ne i}x_j^2}{\sqrt{x_1^2 + \cdots + x_i^2}^{k+2}}
$$
Summing in the divergence will yield $n$ instead of $i$ and $\sum_{i=1}^{n}x_i^2$ on the top, which will cancel with bottom one and will get the original result $\blacksquare$.
A: Without breaking into coordinates, we can just directly evaluate the derivatives using the chain rule:
$$\nabla \frac{r}{|r|^k} = \frac{\nabla \cdot r}{|r|^k} + (\nabla [r^2]^{-k/2}) r$$
The first term is, of course, just $n/|r|^k$.  The choice of writing things in terms of $r^2$ is for simplicity.
$$\nabla [r^2]^{-k/2} = -\frac{k}{2} [r^2]^{-k/2-1} (2r) = -kr |r|^{-k-2}$$
We can directly evaluate the result to get
$$\nabla \frac{r}{|r|^k} = \frac{n}{|r|^k} - \frac{k r^2}{|r|^{k+2}} = \frac{n-k}{|r|^k}$$
The result is just a scalar; there are no bivector terms.
