The gradient of the form $\text{Tr}\left(W(I+X^HAX)^{-1}\right)$ I read a paper and the paper shows the following:
$$\nabla_{B_k} \text{Tr}\left( W_k E_k \right)=-H^H_kR^{-1}H_kB_kE_kW_kE_k,$$
where $E_k=\left( I+B^H_k H^H_kR^{-1}H_kB_k \right)^{-1}$.
Here,
$H_k \in \mathbb{C}^{Q \times P}$,
$B_k \in \mathbb{C}^{P \times Q}$,
$W_k \in \mathbb{C}^{Q \times Q}$
$R^{-1}$ is $Q \times Q$ hermitian matrix that is independent of $B_k$. However, I try to derive the gradient myself and instead I obtain
$$\nabla_{B_k} \text{Tr}\left( W_k E_k \right)=-E_kH^H_kR^{-1}H_kB_kE_kW_k.$$
My approach is the following. I use the fact that:
$$E_k=E_k^H$$
$$\partial \text{Tr} \left( W_kE_k^H\right)=W_k^H,$$
$$\partial J^{-1}=-J^{-1} \frac{\partial J}{\partial B_k} J^{-1},$$
$$J = I+B^H_k H^H_kR^{-1}H_kB_k,$$
I apply the chain rule and thus obtain
$$\frac{\partial \text{Tr} \left( W_kE_k\right)}{\partial B_k}=
-W_k^H E_k B^H_k H^H_kR^{-1}H_k E_k.$$
Am I doing something wrong somewhere? Can someone kindly help me out?
 A: Let's reduce the clutter by dropping the $k$ subscripts.
Then define the hermitian matrices
$$\eqalign{
G &= H^HR^{-1}H \;&=\; G^H \\
F &= I+B^H GB \;&=\; F^H \\
E &= F^{-1} \;&=\; E^H  \\
dE &= -E\,dF\,E \\
}$$
Let's also use a colon as a convenient product notation for the trace, i.e.
$$\eqalign{
A:B &= {\rm Tr}(AB^T) \;=\; \sum_{i=1}^m \sum_{j=1}^n A_{ij} B_{ij} \\
A^*:A &= \big\|A\big\|^2_F \\
}$$
Write the objective function using the above notation.
Then calculate its differential and gradient $\big($treating $B^*$ as independent of $B\big)$.
[NB: After re-reading the comments, I conjugated the function]
$$\eqalign{
\phi^* &= W^*:E \\
d\phi^* &= W^*:dE \\
 &= -W^*:E\,dF\,E \\
 &= -E^*W^*E^*:dF \\
 &= -E^*W^*E^*:B^HG\,dB \\
 &= -G^*B^*E^*W^*E^*:dB \\
 &= -(GBEWE)^*:dB \\
\frac{\partial\phi^*}{\partial B}
 &= -(GBEWE)^*\\
}$$
This is the complex conjugate of your report of the result from the paper.
Therefore I assume that the paper actually claims that
$$\eqalign{
\frac{\partial\phi}{\partial B^*} &= -GBEWE \\
}$$
and you simply misread it.
A: Let $f(X) = \operatorname{tr}(W(I+X^HAX)^{-1})$ (correcting  the error in the question title). Use the chain rule through to find the map derivative
$\mathrm d_X  f=\operatorname{tr}\left(W\cdot \mathrm d_X\left[(I+X^HAX)^{-1}\right]\right)=\operatorname{tr}\left(-W\cdot ()^{-1}\cdot d_X\left[()\right] \cdot ()^{-1}\right) = -\operatorname{tr}\left(W\cdot (I+X^HAX)^{-1}\cdot (X^HA\mathrm d_XX+\mathrm (d_XX)^H A X) \cdot (I+X^HAX)^{-1}\right)$
The $\mathrm d_XX$ is the identity map of course and is serving here as a placeholder: that's where an argument $h$ of this linear map gets plugged in. Let's also rename variables $B=X$, $G=A$ and $E=(I+B^HGB)^{-1}$. We now have
$\mathrm d_B  f[h] = -\operatorname{tr}\left(W\cdot E\cdot (B^HGh+h^H G B) \cdot E\right)$
We now use the cyclic, linearity, and transpose properties of trace:
$-\mathrm d_B  f[h] = \operatorname{tr}\left(EWE\cdot (B^HGh+h^H G B)\right)=\operatorname{tr}(EWEB^HG\cdot h)+\operatorname{tr}(EWE\cdot h^HGB)=\operatorname{tr}((EWEB^HG+(GBEWE)^H)\cdot h))=\langle(EWEB^HG)^H+GBEWE,h\rangle_F$
where we also brought in the definition of the Frobenius inner product. Therefore the "gradient" is $-((EWEB^HG)^H+GBEWE)$.
In your case $G^H=G$ and $E^H=E$ so the answer is indeed $-GBE(W^H+W)E$
