Is there a quick way to calculate the derivative of a quantity that uses Einstein's summation convention?

Question

Consider $F_{\mu\nu}=\partial_{\mu}A_\nu-\partial_\nu A_\mu$, I am trying to understand how to fast calculate $$\frac{\partial(F_{\mu\nu}F^{\mu\nu})}{\partial (\partial_\alpha A_\beta)}$$

without expanding the multiplication in terms of $A$. Clearly if we do not have indices, this can be done very quickly with a glance, yet when there are indices multiplication factors come into play we would, in this case get a factor of $4$ before usual derivative.

Is there a way to glance at it and get correct answer immediately?

Answer 1

The product rule (Leibniz rule) applies to functional derivatives, so we have $$ \frac{\partial(F_{\mu\nu}F^{\mu\nu})}{\partial (\partial_\alpha A_\beta)} = 2 F^{\mu \nu} \frac{\partial(F_{\mu\nu})}{\partial (\partial_\alpha A_\beta)} = 2 F^{\mu \nu}\left( \delta^{\alpha}_{\mu} \delta^{\beta}_{\nu} - \delta^{\alpha}_{\nu} \delta^{\beta}_{\mu}\right) = 4 F^{\alpha \beta}. $$ The step where I actually take the derivative of $F_{\mu \nu}$ with respect to $\partial_\alpha A_\beta$ does involve writing out the field strength in terms of the derivatives, but it's less complicated in this version.