The derivative of the determinant of a Kronecker product For an invertible matrix $A$, we have the identity
\begin{align}
\dfrac{\partial \det A}{\partial A} = \det A (A^{-1})^T
\end{align}
where the $T$ denotes the transpose operation.
How does this formula change when considering the Kronecker product of $A$ with some other (invertible) matrix $B$? For example, how does one compute the following:
\begin{align}
\dfrac{\partial \det \left[A\otimes B\right]}{\partial A}
\end{align}

I find this question very interesting. I could not find any reference to this problem in any book. So, here is my derivation for square matrices and I would like to hear other people to know if it is correct.
$A$ being ${n\times n}$ matrix and $B$ being ${p\times p}$ matrix. We know that $$\det[A\otimes B]=(\det{A})^p(\det{B})^n$$
(Notice how the power switches between $n$ and $p$)
Then the derivative is $$\frac{\partial(\det{A})^p(\det{B})^n}{\partial A}=(\det{B})^np(\det A)^{p-1}\det A(A^{-1})^T=p(\det B)^n(\det A)^p(A^{-1})^T$$
To put it "nicer":
$$\frac{\partial\det[A\otimes B]}{\partial A}=p\det[A\otimes B](A^{-1})^T$$
