I need to calculate $\frac{\partial f(X)}{\partial X}$ for $f(X)=tr \big[ C(X^\top BX)^{-1}\big]$. Where $\{X,B,C\}$ are matrices.
I tried to assume it as $g(U)=tr(CU^{-1})$ when $U=X^\top BX$, then using the chain rule $$\frac{\partial f(X)}{\partial X}=tr\Big(\frac{\partial g(U)}{\partial U}. \frac{\partial U}{\partial X_{ij}} \Big)$$.
I think the first part should be $\frac{\partial g(U)}{\partial U}=-U^{-1}CU^{-1}$, but i'm not sure about how to derive the 2nd part.
Define the matrices
$$\eqalign{
M &= X^TBX \cr
P &= M^{-1}CM^{-1} \cr
}$$
and write the function in terms of $M$ and the inner product (denoted by a colon)
$$\eqalign{
f &= C^T:M^{-1} \cr\cr
}$$
Now find the differential and gradient
$$\eqalign{
df &= -C^T:M^{-1}\,dM\,M^{-1} \cr
&= -M^{-T}C^TM^{-T}:dM \cr
&= -P^T:dM \cr
&= -P^T:(dX^TBX+X^TB\,dX) \cr
&= -P^TX^TB^T:dX^T \,\,-\,\, B^TXP^T:dX \cr
&= -\big(BXP + B^TXP^T\big):dX \cr
\cr
\frac{\partial f}{\partial X} &= -\big(BXP + B^TXP^T\big) \cr
&= -BXM^{-1}CM^{-1} - B^TXM^{-T}C^TM^{-T} \cr
\cr
}$$
Knowing the trace equivalent of the inner product
$$\eqalign{
A:B &= {\rm tr}(A^TB) \cr
}$$
can be useful for rearranging terms, e.g.
$$\eqalign{
A:BC &= B^TA:C \cr&= AC^T:B \cr&= etc.
}$$