Derivatives with upper and lower indices I'm studying classical and quantum field theory, but evaluating derivatives of fields (scalar and/or vector) described with upper and lower indices is somewhat new to me. I'm trying to evaluate 
$$\frac{\partial}{\partial A_\mu} (F_{\mu \nu}) \text{,} \tag{1}$$ 
where $$F_{\mu \nu}= \partial_{\mu} A_{\nu}- \partial_{\nu} A_{\mu}.\tag{2}$$ 
Just for confirmation, is the result of the above partial derivative just simply $- \partial_{\nu}$? 
I'm trying to use the Euler-Lagrange equation of classical field theory to derive the equations of motion for a Lagrangian density that describes an abelian Higgs model with given potential $- \lambda(\phi^{\dagger} \phi - a^2)^2$ for $\lambda, a$ arbitrary constants and $\phi$ a complex-valued scalar field, but the evaluation of derivatives with upper and lower indices makes it a lot messier. Is there a better or quicker method? 
 A: First, you should be careful with your choice of indices. What you have written in Eq. 1 implies a summation over $\mu$ that I don't think you actually want. It is true that
\begin{equation}
\frac{\partial F_{\mu\nu}}{\partial A_{\sigma}}=0,
\end{equation}
but that is just because $F_{\mu\nu}$ depends on the derivatives of $A_{\mu}$ and not $A_{\mu}$ 
explicitly. Let's work out a more full example to see what's going on with the upper and lower indices. Suppose we have the Lagrangian
\begin{equation}
\mathcal{L}=-\frac{1}{4}F_{\mu\nu}^2.
\end{equation}
Now, we want to calculate the Euler-Lagrange equations with respect to the gauge field $A_{\mu}$:
\begin{equation}
\partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}A_{\nu})}
=\frac{\partial\mathcal{L}}{\partial A_{\nu}}.
\end{equation}
Recall that indices are raised and lowered using the metric (in this example Minkowski metric $\eta$)
\begin{equation}
A^{\mu} = \eta^{\mu\nu}A_{\nu}.
\end{equation}
Let's start by rewriting the Lagrangian
\begin{equation}
\begin{split}
\mathcal{L}&=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}=-\frac{1}{4}\eta^{\mu\sigma}\eta^{\nu\rho}F_{\mu\nu}F_{\sigma\rho} \\
&= -\frac{1}{4}\eta^{\mu\sigma}\eta^{\nu\rho}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})
(\partial_{\sigma}A_{\rho}-\partial_{\rho}A_{\sigma})
\end{split}
\end{equation}
Now, we are already using $\mu,\nu,\rho$ and $\sigma$ as dummy summation indices, so it would be unwise of us to reuse any of these when computing the Euler-Lagrange equations. We compute
\begin{equation}
\begin{split}
\frac{\partial\mathcal{L}}{\partial(\partial_{\kappa}A_{\lambda})} &=-\frac{1}{4}\eta^{\mu\sigma}\eta^{\nu\sigma}\left[(\delta^{\kappa}_{\mu}\delta^{\lambda}_{\nu} - \delta^{\lambda}_{\mu}\delta^{\kappa}_{\nu})(\partial_{\sigma}A_{\rho}-\partial_{\rho}A_{\sigma}) \right. \\
 &\left.+ (\delta^{\kappa}_{\sigma}\delta^{\lambda}_{\rho} - \delta^{\lambda}_{\sigma}\delta^{\kappa}_{\rho})(\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}) \right] \\
&= -\frac{1}{4}\left[(\eta^{\kappa\sigma}\eta^{\lambda\rho}-\eta^{\kappa\rho}\eta^{\lambda\sigma})(\partial_{\sigma}A_{\rho}-\partial_{\rho}A_{\sigma}) \right. \\
& + \left. (\eta^{\mu\kappa}\eta^{\nu\lambda}-\eta^{\mu\lambda}\eta^{\nu\kappa})
(\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}) \right] \\
&= -(\partial^{\kappa}A^{\lambda} - \partial^{\lambda}A^{\kappa}) \\
&= -F^{\kappa\lambda}
\end{split}
\end{equation}
And again
\begin{equation}
\frac{\partial\mathcal{L}}{\partial A_{\lambda}} = 0,
\end{equation}
since $F$ has no explicit dependence on $A$. Thus
\begin{equation}
\partial_{\kappa}F^{\kappa\lambda}=0.
\end{equation}
