What's $\frac{\partial}{\partial X} f(X\otimes A)$?

Question

Based on the answer to this question, I wonder how, using the differential notation, one finds $\frac{\partial}{\partial X} f(X\otimes A)$?

Assume that $X,A$ are positive definite matrices, and we know what is $\frac{\partial}{\partial U} f(U)$.

Edit: I have in mind this specific $f(U)=d^\intercal U d$

My try: I'll rewrite f as $f(X)=d^\intercal (X\otimes A)d$. We know that if $d=vec(D)$, then $(X\otimes A)d=vec(ADX^\intercal)$. Therefore, $d^\intercal(X\otimes A)d=d^\intercal vec(ADX^\intercal)=Tr(D^\intercal ADX^\intercal)$

so, we have $f=D:ADX^\intercal$, then $df=D:AD \ dX^\intercal=(AD)^\intercal D : (dX)^\intercal$.

Now since f is scalar, we have $df=D^\intercal AD \ dX$ which produces

$$\frac{\partial }{\partial X} f =D^\intercal AD$$

Answer 1

Thanks to Greg, and the answers in the linked questions to this one (see section above, on the right)

I'll rewrite f as $f(X)=d^\intercal (X\otimes A)d$. We know that if $d=vec(D)$, then $(X\otimes A)d=vec(ADX^\intercal)$. Therefore, $d^\intercal(X\otimes A)d=d^\intercal vec(ADX^\intercal)=Tr(D^\intercal ADX^\intercal)$

so, we have $f=D:ADX^\intercal$, then $df=D:AD \ dX^\intercal=(AD)^\intercal D : (dX)^\intercal$.

Now since f is scalar, we have $df=D^\intercal AD \ dX$ which produces

$$\frac{\partial }{\partial X} f =D^\intercal AD$$