How do I calculate the derivative
$\frac{\partial}{\partial\boldsymbol{S}_{1}}\text{tr}\left(\boldsymbol{S}_{i}^{H}\left(\boldsymbol{S}_{1}\boldsymbol{A}_{i1}\boldsymbol{S}_{1}^{H}+\sum_{j\neq i,1}\boldsymbol{S}_{j}\boldsymbol{A}_{ij}\boldsymbol{S}_{j}^{H}\right)^{-1}\boldsymbol{S}_{i}+\boldsymbol{A}_{ii}^{-1}\right)^{-1}$
with $\boldsymbol{S}_{i}$ being $N\times M$ matrices $\forall i$, $N>M$.
One possible approach I am aware for is using the Woodbury identity and notice that $\text{tr}\left(\boldsymbol{S}_{i}^{H}\left(\boldsymbol{S}_{1}\boldsymbol{A}_{i1}\boldsymbol{S}_{1}^{H}+\sum_{j\neq i,1}\boldsymbol{S}_{j}\boldsymbol{A}_{ij}\boldsymbol{S}_{j}^{H}\right)^{-1}\boldsymbol{S}_{i}+\boldsymbol{A}_{ii}^{-1}\right)^{-1}=\text{tr}\left(\boldsymbol{A}_{ii}-\boldsymbol{A}_{ii}\boldsymbol{S}_{i}^{H}\left[\sum_{j}\boldsymbol{S}_{j}\boldsymbol{A}_{ij}\boldsymbol{S}_{j}^{H}\right]^{-1}\boldsymbol{S}_{i}\boldsymbol{A}_{ii}\right)$
and then operate the derivative on the rhs. However, I was wondering if there is a "brute-force" way to operate on the original expression?