The electromagnetic action is given in the language of differential forms by $$S[A]=-\frac{1}{4}\int F\wedge \star F$$ The variation of the electromagnetic action $S$ gives us Maxwell's equations $$d\star F=0.$$
How do you take the variation $\delta S = S[A+\delta A]-S[A]$ of the above action $S$ to obtain Maxwell's equations?
Here we only consider the abelian case. Let $A\to A+\delta A$, and $F\to d(A+\delta A)=dA+d\delta A$.The action becomes
$$S[A+\delta A]=-\frac{1}{4}\int (dA+d\delta A)\wedge\star(dA+d\delta A).$$
So up to terms linear in $\delta A$, $$S[A+\delta A]-S[A]=\frac{1}{2}\int \delta A\wedge d(\star F)+\mathcal{O}(\delta A^2),$$ where we integrated by parts and used the symmetric property of the inner product on $p$ form $(\alpha,\beta)=\int\alpha\wedge\star\beta=\int \beta\wedge \star\alpha=(\beta,\alpha).$
Now $\frac{\delta S}{\delta A}=0$ gives you $d\star F=0$ as $\delta A$ is arbitrary.
