Find a gradient without going into components. Let $ f(\vec{x}) = \frac{1}{| \vec{x} -  \vec{a} |} $.
Find $\nabla f $.

I know how to do this by first splitting it into components and then taking the partials in each direction, to get $\frac{-\vec{x} +  \vec{a}}{| \vec{x} -  \vec{a} |^3} $
My question is how to get this result directly in vector algebra, without going into components.
 A: To start, write
$$
\nabla \frac{1}{|\vec{x}-\vec{a}|}=\nabla[(\vec{x}-\vec{a})\cdot(\vec{x}-\vec{a})]^{-1/2}
=-\frac{1}{2}[(\vec{x}-\vec{a})\cdot(\vec{x}-\vec{a})]^{-3/2}\nabla[(\vec{x}-\vec{a})\cdot(\vec{x}-\vec{a})]
$$
Next, use the (ugly!) gradient-of-scalar-product identity
$$
\nabla(\vec{u}\cdot\vec{v})=(\vec{u}\cdot\nabla)\vec{v}+(\vec{v}\cdot\nabla)\vec{u}+\vec{u}\times(\nabla\times\vec{v})+\vec{v}\times(\nabla\times\vec{u}),
$$
which you can find in Wikipedia's "Del" article, to compute
$$
\begin{align}
\nabla[(\vec{x}-\vec{a})\cdot(\vec{x}-\vec{a})]&=2[(\vec{x}-\vec{a})\cdot\nabla](\vec{x}-\vec{a})+2(\vec{x}-\vec{a})\times(\nabla\times(\vec{x}-\vec{a}))\\&=2[(\vec{x}-\vec{a})\cdot\nabla]\vec{x}+2(\vec{x}-\vec{a})\times(\nabla\times\vec{x})
\end{align}
$$
Note that in the last line we dropped terms involving $\nabla$ acting on $\vec{a}$ because they are zero.
Now, note that $[(\vec{x}-\vec{a})\cdot\nabla]\vec{x}=\vec{x}-\vec{a}$ and $\nabla\times\vec{x} = 0$. These identities are easily proved using index notation:
The first is $(x_j-a_j)\nabla_j x_i=(x_j-a_j)\delta_{ji}=x_i-a_i$ and the second is $\epsilon_{ijk}\nabla_j x_k=\epsilon_{ijk}\delta_{jk}=\epsilon_{ijj}=0$. Note that both need the obvious identity $\nabla_i x_j=\delta_{ij}$.
Put this all together to get
$$
\nabla \frac{1}{|\vec{x}-\vec{a}|}=-[(\vec{x}-\vec{a})\cdot(\vec{x}-\vec{a})]^{-3/2}(\vec{x}-\vec{a})=-\frac{\vec{x}-\vec{a}}{|\vec{x}-\vec{a}|^3}
$$
The whole thing can be more easily done using index notation throughout:
$$
\begin{align}
\nabla_i\frac{1}{|\vec{x}-\vec{a}|}&=\nabla_i[(x_j-a_j)(x_j-a_j)]^{-1/2}\\
&=-\frac{1}{2}[(x_j-a_j)(x_j-a_j)]^{-3/2}\nabla_i[(x_k-a_k)(x_k-a_k)]\\
&=-[(x_j-a_j)(x_j-a_j)]^{-3/2}(x_k-a_k)\nabla_i(x_k-a_k)\\&=-[(x_j-a_j)(x_j-a_j)]^{-3/2}(x_k-a_k)\delta_{ik}\\
&=-\frac{x_i-a_i}{|\vec{x}-\vec{a}|^3}
\end{align}
$$
