Understanding why $\frac{\partial (F_{\mu \nu} F^{\mu \nu})}{\partial (\partial_\lambda A_\beta)}=4 F^{\lambda \beta}$ in Maxwell's Equations [duplicate]

Question

In trying to derive Maxwell's equations from

$$S=\int d^4 x\left(-\frac 1 4 F_{\mu \nu}F^{\mu \nu}\right)$$

Where

$$F_{\mu \nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$$

I'm trying to show that

$$\frac{\partial (F_{\mu \nu} F^{\mu \nu})}{\partial (\partial_\lambda A_\beta)}=4 F^{\mu \nu}$$

But there seems to be something wrong with how I'm going about doing this. My work currently:

$$\frac{\partial F_{\mu \nu}}{\partial(\partial_\lambda A_\beta)} = \frac{\partial F^{\mu \nu}}{\partial(\partial_\lambda A_\beta)}=\delta_{\lambda \mu}\delta_{\beta \nu}-\delta_{\lambda \nu}\delta_{\beta \mu}$$

Where $\delta$ is the Kronecker-Delta function. From this we find that

$$\frac{\partial}{\partial(\partial_\lambda A_\beta)}\left(F_{\mu \nu}F^{\mu \nu}\right)=(F^{\mu \nu}(\delta_{\lambda \mu}\delta_{\beta \nu}-\delta_{\lambda \nu}\delta_{\beta \mu}))+(F_{\mu \nu}(\delta_{\lambda \mu}\delta_{\beta \nu}-\delta_{\lambda \nu}\delta_{\beta \mu}))=(F^{\lambda \beta}-F^{\beta \lambda})+(F_{\lambda \beta}-F_{\beta \lambda})$$

Asserting that $F^{\mu \nu} = -F^{\nu \mu}$ and renaming the dummy indices we get

$$\frac{\partial}{\partial(\partial_\lambda A_\beta)}\left(F_{\mu \nu}F^{\mu \nu}\right)=2(F^{\mu \nu})+2(F_{\mu \nu})$$

Which is a bit off target. Did I go off course somewhere in my current work or am I just missing a few final steps?

Answer 1

$$ \frac{\partial F_{\mu \nu}}{\partial (\partial_{\lambda} A_{\beta})} = \frac{\partial (\partial_{\mu} A_{\nu})}{\partial (\partial_{\lambda} A_{\beta})} - (\mu \leftrightarrow \nu) = \delta^{\lambda}_{\mu} \delta^{\beta}_{\nu} - (\mu \leftrightarrow \nu )$$. The bidirectional arrow means that the second term is the same as the first except that the indices $\mu$ and $\nu$ are interchanged.

Also when the indices on $F$ are upstairs,

$$ \frac{\partial F^{\mu \nu}}{\partial (\partial_{\lambda} A_{\beta})} = \frac{\partial (\partial^{\mu} A^{\nu})}{\partial (\partial_{\lambda} A_{\beta})} - (\mu \leftrightarrow \nu) = \delta^{\lambda \mu} \delta^{\beta \nu} - (\mu \leftrightarrow \nu )$$.

Now use the Leibniz rule and the properties of the Kronecker delta to get to your answer.