Vector analysis. Del and dot products I am trying to prove that
$$\nabla(\mathbf{A} \cdot \mathbf{B}) = (\mathbf{A} \cdot \nabla)\mathbf{B} + (\mathbf{B} \cdot \nabla)\mathbf{A} + \mathbf{A} \times (\nabla \times \mathbf{B}) + \mathbf{B} \times (\nabla \times \mathbf{A})$$
I've gotten as far as $\nabla(\mathbf{A} \cdot \mathbf{B}) = \nabla A\cdot B+\nabla B\cdot A$, using subscript summation. I don't know how to proceed.
This is part of proving that
$$\frac{Dv}{Dt}=\frac{\partial v}{\partial t}+\nabla(\frac{v^2}{2})-v\times(\nabla\times v)$$
 A: Let's start with the right hand side, we have, considering the $i$th component
\begin{align*}
 &\!\!\! [(A \cdot \nabla) B + (B \cdot \nabla)A + A \times (\nabla \times B) + B \times (\nabla \times A)]_i\\
  &= \delta^{jk}A_j\partial_kB_i + \delta^{jk}B_j\partial_kA_i + \epsilon^{jk}_{\;\;i}A_j\epsilon^{\mu\nu}_{\;\;k}\partial_\mu B_\nu + \epsilon^{jk}_{\;\;i}B_j\epsilon^{\mu\nu}_{\;\;k}\partial_\mu A_\nu\\
  &= \delta^{jk}A_j\partial_kB_i + \delta^{jk}B_j\partial_kA_i  + (\delta^\mu_i\delta^{\nu j} - \delta^\nu_i \delta^{\mu j})(A_j\partial_\mu B_\nu + B_j\partial_\mu A_\nu)\\
  &= \delta^{jk}A_j\partial_kB_i + \delta^{jk}B_j\partial_kA_i + \delta^{jk}(A_j\partial_i B_k + B_j\partial_i A_k) - \delta^{jk}(A_j\partial_k B_i + B_j\partial_k A_i)\\
  &= \delta^{jk}(A_j\partial_i B_k + B_j\partial_i A_k)\\
  &= \partial_i(\delta^{jk} A_j B_k)\\
  &= [\nabla(A \cdot B)]_i
\end{align*}
A: Following your effect of using subscript notation: $\newcommand{\b}{\boldsymbol}$
$$
\nabla (\b{A}\cdot \b{B}) = \nabla_{\b{A}}(\b{A}\cdot \b{B}) + \nabla_{\b{B}}(\b{A}\cdot \b{B}),\tag{1}
$$
where $\nabla_{\b{A}}(\b{A}\cdot \b{B})$ is that $\b{A}$ is differentiated while $\b{B}$ is held constant. We can prove:
$$
\nabla_{\b{A}}(\b{A}\cdot \b{B}) = (\b{B}\cdot \nabla) \b{A} + \b{B}\times(\nabla \times\b{A}).\tag{2}
$$
This is a very geometrical identity in that, for $\b{B}$ fixed, the derviate of $\b{A}$ can be decomposed into the normal derivative and the tangential derivative along $\b{A}$:
\begin{align}
&\nabla_{\b{A}}(\b{A}\cdot \b{B}) = 
\begin{pmatrix}\partial_x A_1 &\partial_x A_2 & \partial_x A_3 
\\ \partial_y A_1& \partial_y A_2 & \partial_y A_3
\\ \partial_z A_1 & \partial_z A_2 & \partial_z A_3
\end{pmatrix}\begin{pmatrix}B_1\\B_2 \\B_3\end{pmatrix}
\\
=& 
\begin{pmatrix}\partial_x A_1 &\partial_y A_1 & \partial_z A_1 
\\ \partial_x A_2& \partial_y A_2 & \partial_z A_2
\\ \partial_x A_3 & \partial_y A_3 & \partial_z A_3
\end{pmatrix}\begin{pmatrix}B_1\\B_2 \\B_3\end{pmatrix}
\\
&+ 
\begin{pmatrix}0 & \partial_x A_2-\partial_y A_1 & \partial_x A_3 -\partial_z A_1 
\\ \partial_y A_1 - \partial_x A_2& 0 & \partial_y A_3 - \partial_z A_2
\\  \partial_z A_1 - \partial_x A_3 & \partial_z A_2-\partial_y A_3 & 0
\end{pmatrix}\begin{pmatrix}B_1\\B_2 \\B_3\end{pmatrix}
\\
=&(\b{B}\cdot \nabla) \b{A} + \b{B}\times(\nabla \times\b{A}),
\end{align}
extraction of the anti-symmetric part, for the cross product(wedge) can be written as the following anti-symmetric matrix form:
$$
\begin{pmatrix}
0 & -\partial_z & \partial_y
\\
\partial_z & 0 & -\partial_x
\\
-\partial_y & \partial_x & 0
\end{pmatrix}
\begin{pmatrix}
A_1\\A_2\\A_3
\end{pmatrix} = \nabla \times \b{A}.
$$
The rest is plugging (2) into (1), and do the same for the other term.
A: I always find for identities like these, its best to start from the RHS and get to the LHS - because its an equality, you can go either way you want. Starting with $\mathbf{A} \times (\nabla \times \mathbf{B})$, we see that as its i-th component is (where $\epsilon_{ijk}$ is the Levi-Civita tensor):
$$    [\mathbf{A} \times (\nabla \times \mathbf{B})]_i = \epsilon_{ijk}A_j[\nabla \times \mathbf{B}]_k = \epsilon_{ijk}\epsilon_{klm}A_j\partial_lB_m$$
where we use the notation 
$$\partial_i := \frac{\partial}{\partial x_i}$$
and hence by the identity 
$$ \epsilon_{kij}\epsilon_{klm} = \delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$$
we get that
$$    [\mathbf{A} \times (\nabla \times \mathbf{B})]_i  =  A_j\partial_iB_j - [(\mathbf{A} \cdot \nabla)\mathbf{B}]_i$$
where we realise that the first term on the RHS is half of what we need to get on the left hand side. Doing similarily to $[\mathbf{B} \times (\nabla \times \mathbf{A})]_i$ gives the result.
