Simplifing formulas using tensor notation Im trying to symplify formulas like: 
$$\operatorname{div}(\operatorname{rot}\vec{F}),\qquad \operatorname{rot}(\operatorname{rot}\vec{F}) $$
or something more strange like:
$$\operatorname{rot}(\vec{r}\operatorname{div}(r^4\operatorname{grad}(r^4)))$$
To do this I want to use the tensor notation and by this I mean using the Einstein convention, Levi-Civita symbol, Kronecker delta and all that. The problem is that I don't understand the rules and what its allowed and what not.
As an example:
$$\operatorname{div}(f( r)\cdot \textbf r)= \partial_i(f( r)\cdot \textbf r)_i = \partial_i(f( r)\cdot x_i)= \partial _if(r)x_i+f(r)\partial_ix_i= f'(r)\frac{x_i}{r}x_i+3f(r)=rf'(r)+3f(r) $$
Any help on how to do this kind of problems or where I can find useful examples? 
I'm working in flat space.
Thanks!
 A: The gradient, divergence, curl, and Laplacian can be written in the following way: 
$$\begin{eqnarray*}
(\mathrm{grad}\, f)_i &=& \partial_i f \\
\mathrm{div}\,{\bf F} &=& \partial_i F_i \\
(\mathrm{rot}\,{\bf F})_i &=& \epsilon_{ijk} \partial_j F_k \\
\Delta &=& \partial^2 = \partial_i\partial_i.
\end{eqnarray*}$$
The Levi-Civita symbol has some key properties. 
For example, it is totally antisymmetric so
$\epsilon_{ijk}=-\epsilon_{jik} = \epsilon_{jki}$. 
In addition it has an important multiplicative property: 
$$\begin{eqnarray*}
\epsilon_{ijk}\epsilon_{ilm} &=& \delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}.
\end{eqnarray*}$$
The Levi-Civita symbol and the Kronecker delta are just numbers, so they commute with derivatives.
Below we give some sample calculations. 
Example 1: 
$$\begin{eqnarray*}
\mathrm{div}(\mathrm{rot}\,{\bf F})
&=& \partial_i (\mathrm{rot}\,{\bf F})_i \\
&=& \partial_i (\epsilon_{ijk} \partial_j F_k) \\
&=& \epsilon_{ijk} \partial_i \partial_j F_k \\
&=& \epsilon_{ijk} \partial_j \partial_i F_k \\
&=& -\epsilon_{jik} \partial_j \partial_i F_k \\
&=& -\epsilon_{ijk} \partial_i \partial_j F_k \\
&=& 0
\end{eqnarray*}$$
Example 2: 
$$\begin{eqnarray*}
[\mathrm{rot}(\mathrm{rot}\,{\bf F})]_i
&=& \epsilon_{ijk} \partial_j (\mathrm{rot}\,{\bf F})_k \\
&=& \epsilon_{ijk} \partial_j (\epsilon_{klm}\partial_l F_m) \\
&=& \epsilon_{ijk} \epsilon_{klm} \partial_j \partial_l F_m \\
&=& \epsilon_{kij} \epsilon_{klm} \partial_j \partial_l F_m \\
&=& (\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}) \partial_j \partial_l F_m \\
&=& (\partial_m \partial_i - \delta_{im} \partial^2)F_m \\
&=& \partial_i(\mathrm{div}\,{\bf F}) - (\Delta{\bf F})_i
\end{eqnarray*}$$
Therefore, 
$$\mathrm{rot}(\mathrm{rot}\,{\bf F}) 
= \mathrm{grad}(\mathrm{div}\,{\bf F}) - \Delta{\bf F}.$$
