Problem Taking a Derivative using Einstein Notation At some point of an electrodynamics exercise, I had to prove:
$$ \nabla \frac{1}{|\textbf{r} - \textbf{r}'|} = - \frac{\textbf{r} - \textbf{r}'}{|\textbf{r} - \textbf{r}'|^3}$$
So, using Einstein Notation, I proceeded as follows:
$$\nabla [ (x_k - x_k')^2]^{-\frac{1}{2}} = \frac{\partial}{\partial x_i} [ (x_k - x_k')^2]^{-\frac{1}{2}} \textbf{e}_i = \ ...$$
and now the part I think I am getting wrong
$$... \ (-1/2) [(x_k - x_k')^2]^{-\frac{3}{2}} \cdot 2\cdot (x_k-x_k') \cdot \frac{\partial x_k}{\partial x_i} \textbf{e}_i = $$
$$- [(x_k - x_k')^2]^{-\frac{3}{2}} (x_k - x_k') \delta_{ki} \textbf{e}_i = $$
$$ - [(x_i - x_i')^2]^{-\frac{3}{2}} (x_i - x_i') \textbf{e}_i = $$
$$- \frac{(x_i - x_i') \textbf{e}_i}{[(x_i - x_i')^2]^{\frac{3}{2}}}$$
which of course is not the quantity of the second-hand side. The quantity derived would be correct if it had a different index on the denominator, but the Kronecker delta reduced the double sum to a single sum. What did I do wrong?
$\textbf{Edit 1: Clarification}$
The operator nabla above is:
$$\nabla = \frac{\partial}{\partial x_i} \textbf{e}_i$$
and the vectors $\textbf{r}$ and $\textbf{r}'$ are respectively:
$$\textbf{r} = x_i \textbf{e}_i \text{ and } \textbf{r}' = x_i' \textbf{e}_i$$
 A: Since $$\partial_i|\mathbf{r}-\mathbf{r}^\prime|^2=\partial_i[(x_k-x_k^\prime)(x_k-x_k^\prime)]=2(x_k-x_k^\prime)\partial_i(x_k-x_k^\prime)=2(x_k-x_k^\prime)\delta_{ik}=2(x_i-x_i^\prime),$$the chain rule gives$$\partial_i|\mathbf{r}-\mathbf{r}^\prime|^{-1}=-\frac12|\mathbf{r}-\mathbf{r}^\prime|^{-3}\partial_i|\mathbf{r}-\mathbf{r}^\prime|^2=-|\mathbf{r}-\mathbf{r}^\prime|^{-3}(x_i-x_i^\prime).$$This is the desired $i$th component.
A: Already when you write $|\mathbf{r}-\mathbf{r'}| = [(x_k-x_k')^2]^{-1/2}$ you get it wrong if you add a summation in front of the expression:
$$
\sum_k [(x_k-x_k')^2]^{-1/2} = \frac{1}{|x_1-x_1'|} + \cdots + \frac{1}{|x_n-x_n'|}.
$$
You need to learn to know when you can apply the Einstein summation convention.

One way to do the calculations:
$$\begin{align}
\nabla \frac{1}{|\mathbf{r} - \mathbf{r}'|} 
&= (\mathbf{e}^i \partial_i) \frac{1}{|\mathbf{r} - \mathbf{r}'|} \\
&= (\mathbf{e}^i \partial_i) \left((\mathbf{r} - \mathbf{r}')^2\right)^{-1/2}
\\
&= \mathbf{e}^i \, (-\frac12)\, \left((\mathbf{r} - \mathbf{r}')^2\right)^{-3/2} \left(2(\mathbf{r} - \mathbf{r}')\cdot\partial_i(\mathbf{r} - \mathbf{r}')\right) \\
&= -\mathbf{e}^i \, |\mathbf{r} - \mathbf{r}'|^{-3} \left((\mathbf{r} - \mathbf{r}')\cdot\mathbf{e}_i\right) \\
&= -\mathbf{e}^i \, |\mathbf{r} - \mathbf{r}'|^{-3} (x_i - x'_i) \\
&= -|\mathbf{r} - \mathbf{r}'|^{-3} (\mathbf{r} - \mathbf{r}') \\
&= -\frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|^{3}}
\end{align}$$
