In the context of special relativity I have this to show that this equation is correct: $$\partial_\mu F^{\mu \nu}=j^\nu$$
To do that I'm trying to take this equation: $$F^{\mu \nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$$ and take the 4-gradient of it: $$\partial_\mu F^{\mu \nu}=\partial_\mu\partial^{\mu}A^{\nu}-\partial_\mu\partial^{\nu}A^{\mu}$$ I'm not sure what to do at this point, I think my main problem is that I see these written here and I'm not sure I know what it means. From what I understand $$\partial_\mu=\bigg({\partial\over \partial t},-\vec\nabla\bigg),$$ but I don't know what $\partial^{\mu}$ or $\partial^{\nu}$ are equal to. What happens to $A^{\nu}$ and $A^{\mu}$ when they are acted on by $\partial^{\mu}$ and $\partial^\nu$, respectively? Is this even the right approach to prove that $\partial_\mu F^{\mu\nu}=j^\nu$?
You can lower and raise indices with the metric, which in the case of special relativity is
$$ \eta^{\mu \nu} = \pmatrix{1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1} \tag{1} $$
As an example
$$ \partial^\mu = \eta^{\mu \nu}\partial_\nu \tag{2} $$
It is then easy to calculate
\begin{eqnarray} \require{cancel} \partial^0 &=& \eta^{0\nu}\partial_\nu = \cancelto{1}{\eta^{00}}\partial_0 = \frac{\partial}{\partial t} \\ \partial^i &=& \eta^{i\nu}\partial_\nu = -\frac{\partial}{\partial x^i} ~~~\mbox{for}~~ i = 1,2,3\tag{3} \end{eqnarray}
Same applies for the field $F^{\mu\nu}$,
$$ F^{\mu\nu} = \eta^{\nu\beta}F^{\mu}_{\;\beta} = \eta^{\mu\alpha} \eta^{\nu\beta}F_{\alpha\beta} $$