I have some problems with calculating the derivative of complex matrix functions which involves the chain rule.
1.) I found the following matrix derivative in multiple papers: $$ d\mathbf{X} \log\det\left(\mathbf{I} + \mathbf{BXB}^H\right) = \text{tr}\left( \mathbf{B}^H \left( \mathbf{I} + \mathbf{BXB}^H \right)^{-1} \mathbf{B} d \mathbf{X}\right)$$
They obviously used the following rule: $$d \mathbf{X} \log\det(\mathbf{X}) = \text{tr}(\mathbf{X}^{-1}d\mathbf{X})$$ [see e.g. The Matrix Cookbook eq. 43, p.8] and additionally they must have used the chain rule. But I don't understand why it is: $$ \mathbf{B}^H \left( \mathbf{I} + \mathbf{BXB}^H \right)^{-1} \mathbf{B} $$ and not (exchange of the conjugate transpose $(\cdot)^H$outside of $(\cdot)^{-1}$): $$ \mathbf{B} \left( \mathbf{I} + \mathbf{BXB}^H \right)^{-1} \mathbf{B}^H $$
because $$ d\mathbf{X} (\mathbf{I} + \mathbf{BXB}^H) = \mathbf{B}d\mathbf{XB}^H$$ Can someone explain this result to me?
2.) My overall goal is to compute a derivative of the following form: $$ d\mathbf{X} \log\det\left(\mathbf{I} + \mathbf{BXB}^H + \mathbf{CXC}^H\right) $$ Therefore I would like to understand how to compute this derivatives.
Thank you in advance, Enzo.
Reference: https://www.math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf
Denote by $f$ the function $$ f(X) = \def\I{\mathord{\rm Id}}\I + BXB^*. $$ We have, as you write correctly $$ f'(X)H = BHB^* $$ Hence, using $$ (\log \det)'(X)H = \def\tr{\operatorname{tr}}\tr(X^{-1}H) $$ we get, using the chain rule \begin{align*} (\log\mathord\det\circ f)'(X)H &= (\log \det)'\bigl(f(X)\bigr)f'(X)H\\ &= \tr\bigl(f(X)^{-1}f'(X)H\bigr)\\ &= \tr\bigl((\I + BXB^*)^{-1}BHB^*\bigr)\\ &= \tr \bigl(B^*(\I + BXB^*)^{-1}BH\bigr) \end{align*} In the last equation we used the property $\tr(AB) = \tr(BA)$ of the trace. (I used $H$ where you used $\partial X$).