How to simplify expressions with del (or nabla) in them? I always find it difficult to simplify expressions or open brackets in expressions that have a 'Del' (or 'Nabla') in them.
For example, how would one go about simplifying this expression?:
$$\nabla\boldsymbol{\cdot}(\phi\nabla\psi)$$
($\phi$ and $\psi$ are both scalar fields)
I need it to become:
$$[\phi\nabla^2\psi + (\nabla\phi)\boldsymbol{\cdot}(\nabla\psi)]$$
I would also love to know how to simplify those standard equations mentioned in Griffiths (for example - the expansion of the 'curl of the curl' of a vector field)

The only method I know is to

*

*find out every single term in the expression (in terms of $a_x$, $a_y$ etc.)

*and then cancel out the terms

*and then find patterns and regroup the terms in the remaining expression


Is there a faster way to approach these 'simplify' (or 'expand') problems? Maybe there are some tricks or formulas that I am unaware of (maybe something analogous to the uv-rule for differentiating the product of two functions in simple calculus)
$$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$
I understand that the uv-rule seems to work on my original expression. But I would still love some sort of formalization. The problem I have is that, in simple calculus, multiplying two functions does not have two meanings.
With Nabla however, I have two choices - Dot product and Cross product.
And I also have three choices for differentiation - Gradient, Divergence and Curl
To explain my concern better, try answering what would have been the simplification if the original expression was -
$$\nabla \times (\phi\nabla\psi)$$
or maybe
$$\nabla(v\boldsymbol{\cdot}\nabla\psi)$$
where $v$ is a vector-field
For the analogy, these three questions become the same question -

"Differentiation of something multiplied by the differential of something else"

 A: You are computing the divergence of the vector field $\left(\phi \frac{\partial \psi}{\partial x_i}\right)_{i=1,\cdots,n}$, so you just get
$$
\sum_{i=1}^n \frac{\partial}{\partial x_i} \left(\phi \frac{\partial \psi}{\partial x_i}\right)
$$
using the product rule you simply get
$$
\sum_{i=1}^n \left(\frac{\partial \phi}{\partial x_i} \frac{\partial \psi}{\partial x_i} + \phi \frac{\partial^2 \psi}{\partial x_i^2}\right) = \nabla \phi \cdot \nabla \psi + \phi \nabla^2\psi
$$
once you know the result, you can "build" some mnemonic related to the product rule, but you still need to know what first and second order operators you must use.
A: There are many identities in vector calculus that can be referred to simplify such expressions.
Using $\nabla\cdot(\phi\mathbf A)=\phi\nabla\cdot\mathbf A+(\nabla\phi)\cdot\mathbf A$, which looks analogous to the product rule of differentiation, you get
$$\nabla.(\phi\vec\nabla\psi)=\phi\nabla^2\psi + (\vec\nabla\phi).(\vec\nabla\psi)$$
EDIT:
Consider ${\color{red}{\mathbf C}}\times(\mathbf A\times\mathbf B)=\mathbf A(\mathbf {\color{red}{\mathbf C}}\cdot\mathbf B)-\mathbf B(\mathbf {\color{red}{\mathbf C}}\cdot\mathbf A)$ and ${\color{red}{\mathbf\nabla}} \times (\mathbf{A} \times \mathbf{B})
     \ =\  \mathbf{A}({\color{red}{\mathbf\nabla}} {\cdot} \mathbf{B}) \,-\, \mathbf{B}({\color{red}{\mathbf\nabla}} {\cdot} \mathbf{A}) 
           \,+\, (\mathbf{B} {\cdot} \nabla) \mathbf{A} \,-\, (\mathbf{A} {\cdot} \nabla) \mathbf{B}$ .
Where is the analogy? I think after deriving a few formulae listed in the link attached, you could tell apart where the analogy works and where it doesn't.
A: I'll expand what I've commented above here.
Using suffix notation and summation convention (since we are working with $\mathbb{R}^n$ there is no need to distinguish upstairs and downstairs indices, so just write everything downstairs), you can get, for example,
\begin{align*}\require{color}
[\nabla\times(\mathbf{A}\times\mathbf{B})]_i
&=\epsilon_{ijk}\partial_j(\mathbf{A}\times\mathbf{B})_k\\
&=\epsilon_{ijk}\epsilon_{k\ell m}\partial_j A_\ell B_m\\
&=(\delta_{i\ell}\delta_{jm}-\delta_{im}\delta_{j\ell})\partial_j A_\ell B_m\\
&={\color{red}\partial_j A_i B_j}-{\color{blue}\partial_j A_j B_i}\\
\end{align*}
Note that, by convention, $\partial$ acts on everything to its right.  Whereas in the case of $\mathbf{C}\times(\mathbf{A}\times\mathbf{B})$ we would just "take out" the $A_i$ and $B_i$ from the red and blue term respectively, here we can't move them in front of the differential operator without paying for the noncommutativity:
\begin{align*}
{\color{brown}\partial_j A_i} B_j&={\color{brown}A_i\partial_j}B_j+({\color{brown}\partial_jA_i})B_j\\
&=A_i\partial_jB_j+B_j\partial_jA_i\\
\partial_j A_j B_i&=B_i\partial_jA_j+A_j\partial_jB_i.
\end{align*}
So
$$
\nabla\times(\mathbf{A}\times\mathbf{B})=\mathbf{A}(\nabla\cdot\mathbf{B})-\mathbf{B}(\nabla\cdot\mathbf{A})+(\mathbf{B}\cdot\nabla)\mathbf{A}-(\mathbf{A}\cdot\nabla)\mathbf{B}
$$
and you can see the additional terms are precisely what we get from moving something behind a $\nabla$ to in front of that $\nabla$.
Now we have done the calculation, you might reasonably ask the question: Can one immediately get from
$$
\mathbf{C}\times(\mathbf{A}\times\mathbf{B})=\mathbf{A}(\mathbf{C}\cdot\mathbf{B})-\mathbf{B}(\mathbf{C}\cdot\mathbf{A})
$$
to a formula for $\nabla\times(\mathbf{A}\times\mathbf{B})$?  To start with, we check the naive substitution still gives terms that make sense (i.e., it doesn't leave dangling $\nabla$).  Then we see the formula involves pushing $\mathbf{A}$ (or $\mathbf{B}$ in the second term) in front of $\mathbf{C}$, so we need to compensate that by having something from $\nabla\mathbf{A}$ (or $\nabla\mathbf{B}$).  So the formula has to read something like
$$
\mathbf{A}(\nabla\cdot\mathbf{B})-\mathbf{B}(\nabla\cdot\mathbf{A})+(\nabla\mathbf{A})\ast\mathbf{B}-(\nabla\mathbf{B})\ast\mathbf{A}
$$
where $\ast$ does some contraction between the rank-2 tensor and the vector.  Now it is not hard to see in $(\nabla\mathbf{A})\ast\mathbf{B}$ the $\mathbf{B}$ has to contract with the $\nabla$ rather than $\mathbf{A}$ (because the term we are correcting has that), hence we obtain
$$
\mathbf{A}(\nabla\cdot\mathbf{B})-\mathbf{B}(\nabla\cdot\mathbf{A})+(\mathbf{B}\cdot\nabla)\mathbf{A}-(\mathbf{A}\cdot\nabla)\mathbf{B}.
$$
Curl of  a curl:  Similarly,
$$
\mathbf{C}\times(\mathbf{A}\times\mathbf{B})=\mathbf{A}(\mathbf{C}\cdot\mathbf{B})-(\mathbf{C}\cdot\mathbf{A})\mathbf{B}
$$
The right hand side still make sense when $\mathbf{A}=\mathbf{C}=\nabla$.
$$
\nabla\times(\nabla\times\mathbf{B})=\nabla(\nabla\cdot\mathbf{B})-(\nabla\cdot\nabla)\mathbf{B}+\color{red}\text{correction}
$$
We note that we didn't push any vector field pass a $\nabla$, so there are no correction terms.  (We did change the order of $\mathbf{C}$ and $\mathbf{A}$ in $\mathbf{A}(\mathbf{C}\cdot\mathbf{B})$ but they are both the differential operator $\nabla$ so the symmetry of partial derivatives means there are no correction term).
However, I'd seriously advice against doing this eyeballing for anything more complicated.  To see why, think about $(\mathbf{A}\times\nabla)\times\mathbf{B}=\mathbf{A}\cdot\nabla\mathbf{B}-\mathbf{A}(\nabla\cdot\mathbf{B})=\mathbf{A}\times(\nabla\times\mathbf{B})+(\mathbf{A}\cdot\nabla)\mathbf{B}-\mathbf{A}(\nabla\cdot\mathbf{B})$.
