I am trying to compute the derivative of $F(X)=\log \vert C'\Sigma^{-1}C \vert$, where $C = C(X) = (J_1 + J_2 \otimes X)$. All matrices are real, $\Sigma$ is positive definite, $C$ is invertible for every $X$. Here's where I get stuck:
I compute the differential to be
\begin{align} dF &=\mathrm{tr} \left[(C'\Sigma^{-1}C)^{-1}(C'\Sigma^{-1} dC + (dC)'\Sigma^{-1}C)\right] \\ &= 2\mathrm{tr} \left[(C'\Sigma^{-1}C)^{-1}(C'\Sigma^{-1}) dC \right] \\ &= 2\mathrm{tr} \left[(C'\Sigma^{-1}C)^{-1}(C'\Sigma^{-1}) (J_2 \otimes dX) \right], \end{align}
but where do I go from here? I can't see how to isolate $dX$ on the right hand side. Maybe there is another path I should be taking. Or maybe there is a rule to read the derivative directly from this form of differential.
Writing your differential in terms of the Frobenius Inner Product yields $$\eqalign{ df &= 2\Sigma^{-1}C(C^T\Sigma^{-1}C)^{-1} : J_2\otimes dX \cr &= M : J_2\otimes dX \cr }$$ Assume that we can find a Kronecker Product Decomposition for the matrix on the LHS, i.e. $$\eqalign{ M &= A\otimes B \cr }$$ where the shape of A is similar to J, and B to X.
Now apply the Kronecker-Frobenius mixed product rule $$\eqalign{ df &= (A\otimes B):(J_2\otimes dX) \cr &= (A:J_2) \otimes (B:dX) \cr &= (A:J_2)B:dX \cr }$$ Since $df=\big(\frac{\partial f}{\partial X}:dX\big),\,$ the gradient must be $$\eqalign{ \frac{\partial f}{\partial X} &= (J_2:A)\,B \cr }$$ If you cannot find a single Kronecker product that equals M, you can always find a sum of such products $$\eqalign{ M &= \sum_k A_k\otimes B_k \cr \frac{\partial f}{\partial X} &= \sum_k (J_2:A_k)\,B_k \cr }$$