Lagrangian for the Maxwell equations In his book 'Classical Electrodynamics' Kurt Lechner wants to find a Lagrangian $\mathcal{L}$ so that the Euler Lagrange equations
$$\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)}-\frac{\partial\mathcal{L}}{\partial A_\nu}=0$$
give rise to the Maxwell equations
$$\partial_\mu F^{\mu\nu}=j^\nu.$$
He explains heuristically why it is of the form $\mathcal{L}=\mathcal{L}_1+\mathcal{L}_2$ with
$\mathcal{L}_1\propto F^{\mu\nu}F_{\mu\nu}$ and $\mathcal{L}_2\propto A_\mu j^\mu$ considering gauge and Lorentz invariance. I think I got this part figured out.
But next he shows that the above $\mathcal{L}$ really gives rise to the Maxwell equations and here is where I get lost:
He sets $\mathcal{L}_1 =-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$ to get the normalizations right and then he consideres the variation of $\mathcal{L}_1$ under an infintesimal variation of $\partial A$:
\begin{equation}
\delta\mathcal{L}_1=-\frac{1}{2}F^{\mu\nu}\delta F_{\mu\nu}=-\frac{1}{2}F^{\mu\nu}\left(\delta\partial_\mu A_\nu - \delta\partial_\nu A_\mu\right)=-F^{\mu\nu}\delta\left(\partial_\mu A_\nu\right).
\end{equation}
Can someone please explain the steps he made in each of the equalities in the last equation?
 A: Let
$$-\frac{1}{2}F^{\mu\nu}\left(\delta\partial_\mu A_\nu - \delta\partial_\nu A\mu\right)=-\frac{1}{2}F^{\mu\nu}\delta\partial_\mu A_\nu+\frac{1}{2}F^{\mu\nu}\delta\partial_\nu A\mu$$
Interchange the dummy index $\mu$ and $\nu$ in the second term of the equation.You will get
$$-\frac{1}{2}F^{\mu\nu}\delta\partial_\mu A_\nu+\frac{1}{2}F^{\nu\mu}\delta\partial_\mu A\nu$$
Since $F^{\mu \nu}=-F^{\nu \mu}$ ($F^{\mu \nu}$ is Antisymmetric)
we can write
$$-\frac{1}{2}F^{\mu\nu}\delta\partial_\mu A_\nu+\frac{1}{2}F^{\nu\mu}\delta\partial_\mu A\nu=-\frac{1}{2}F^{\mu\nu}\delta\partial_\mu A_\nu-\frac{1}{2}F^{\mu\nu}\delta\partial_\mu A\nu=-F^{\mu\nu}\delta\left(\partial_\mu A_\nu\right)$$
A: Schematically, $\delta(-\frac{1}{4}F^2)=-\frac{1}{4}(2F)\delta{F}$.
Again, schematically, note that $$\delta f(x)=\tilde{f}(x)-f(x), \hspace{5mm} \partial f(x)=(f(x+dx)-f(x))/dx$$
This means, for example-$$\delta(\partial A)=\frac{1}{dx}\delta(A(x+dx)-A(x))$$
$$=\frac{1}{dx}\bigg((\tilde{A}(x+dx)-\tilde{A}(x))-(A(x+dx)-A(x))\bigg)$$
$$=\frac{1}{dx}\bigg(\delta A(x+dx)-\delta A(x)\bigg)=\partial(\delta A)$$
i.e. $\delta$ and $\partial$ commute. This step isn't really necessary here, but would be essential if you were to carry out the computation in all it's mathematical glory. Finally, note that $$F^{\mu\nu}\delta(\partial_\mu A_\nu)=F^{\nu\mu}\delta(\partial_\nu A_\mu)=-F^{\mu\nu}\delta(\partial_\nu A_\mu)$$
Where I have first switched the dummy indices,  and then used $F^{\mu\nu}=-F^{\nu\mu}$.
The result now follows if you subtract the quantity on the RHS on both sides of the last equation.
