How can we simplify the differential form in the expanded expression $$ d [ A, B]=? $$ Suppose $A$ is $p$-form, and $B$ is $q$-form, and $d$ is exterior derivative.
My attempt: Generally I get $$d [ A, B]=d (AB-BA)= (d A) B +(-1)^p A (d B)- (d B) A - (-1)^{q} B (d A)$$
if both $p, q$ are even, we have $d [ A, B]=d (AB-BA)= (d A) B +A (d B)- (d B) A - B (d A)=[(d A), B ]+[A, (d B)],$ thus $$d [ A, B]=[(d A), B ]+[A, (d B)]$$ when $p, q$ are even integers.
Do we have a simpler/simple expression for general $p,q$? And how can we refer the $d [ A, B]=?$ in general in terms of $A,B$, $dA$, $dB$?
Perhaps the commutator in question is the graded commutator, i.e., $[A,B]=AB-(-1)^{pq}BA$. Because in this case, the formula simplifies to $$d[A,B]=[dA,B]+(-1)^p[A,dB],$$ which seems nicer than anything you get with the (non-graded) commutator.