Index notation for vector calculus proof I’d like to prove that $\nabla v \cdot \nabla w = \frac{1}{2} \Big(\nabla^2(vw) - v\nabla^2 w -w\nabla^2 v\Big)$. I’ve attempted to use index notation, but I am unsure of how to rely on the chain rule to obtain the result. I am unable to intuitively see where the factor of half comes from as well. I seek your help! Thank you!
 A: Generally it's easier to prove these identities starting from the longer side (the right side in this case).  Also, make sure you remember the product rule:
$$\begin{align*}
\nabla^2(vw) - v\nabla^2 w -w\nabla^2 v &= \partial_i^2(vw) - v\partial_i^2w - w\partial_i^2v \\
&= \partial_i(w\partial_i v + v\partial_i w) - v\partial_i^2w - w\partial_i^2v \\
&= (\partial_iw)(\partial_i v) + w\partial_i^2v + (\partial_iv)(\partial_i w) + v\partial_i^2w - v\partial_i^2w - w\partial_i^2v \\
&= 2(\partial_i v)(\partial_i w) \\
&= 2\nabla v \cdot \nabla w
\end{align*}$$
A: The $\vec\nabla$ operator is such that:  $$\vec\nabla(\vec u)= \sum_i \frac{\partial u_i}{\partial x_i}\vec {\sf e}_i$$
So it may be treated as a pseudo vector  $$\nabla_i( u_i )= \dfrac{\partial~u_i}{\partial x_i~}$$
Then the chain rule is applied as normal for a derivative operator; $$\begin{split}\nabla_i(v_iw_i) &= v_i\,\nabla_i(w_i)+\nabla_i(v_i)\,w_i \\[2ex] \therefore\quad \vec\nabla (\vec v\cdot\vec w) &= \vec v\cdot\vec\nabla(\vec w)+\vec\nabla(\vec v)\cdot\vec w\end{split}$$

So, likewise, start at the index notation for $\vec\nabla^2(\vec v\cdot \vec w)$,
$$\nabla_i^2(v_iw_i) {= \nabla_i(\nabla_i(v_iw_i)) \\ =\nabla_i(v_i\nabla_i(w_i)+\nabla_i(v_i)~w_i) \\ = \nabla_i(v_i\nabla_i(w_i))~+~\nabla_i(\nabla_i(v_i)~w_i)}$$
Now just apply the chain rule again on those terms.
