I would like to better understand the Jacobian with respect to $\mathbf{x}$ of the following quadratic form:
$$ \mathbf{h}(\mathbf{x})^H\Sigma^{-1}\mathbf{h}(\mathbf{x}) $$
where $\mathbf{h}(\mathbf{x}) \in \mathbb{C}^{N}$ and $\Sigma^{-1} \in \mathbb{C}^{N\times N}$ is Hermetian.
Is there a significant difference if $\mathbf{x} \in \mathbb{R}^M$ versus $\mathbf{x} \in \mathbb{C}^M$?
Here's what I've looked at:
$$ \mathbf{h}(\mathbf{x} + d\mathbf{x}) \approx \mathbf{h}(\mathbf{x}) + \mathbf{h}'(\mathbf{x})d\mathbf{x} $$
and
$$ \lim_{|d\mathbf{x}| \to 0} \frac{\left(\mathbf{h}(\mathbf{x}) + \mathbf{h}'(\mathbf{x})d\mathbf{x}\right)^H\Sigma^{-1} \left(\mathbf{h}(\mathbf{x}) + \mathbf{h}'(\mathbf{x})d\mathbf{x}\right) - \mathbf{h}(\mathbf{x})^H\Sigma^{-1}\mathbf{h}(\mathbf{x})}{|d\mathbf{x}|} \\ \Rightarrow \lim_{|d\mathbf{x}| \to 0} \frac{d\mathbf{x}^H\mathbf{h}'(\mathbf{x})^H\Sigma^{-1}\mathbf{h}(\mathbf{x}) + \mathbf{h}(\mathbf{x})^H\Sigma^{-1}\mathbf{h}'(\mathbf{x})d\mathbf{x} + d\mathbf{x}^H\mathbf{h}'(\mathbf{x})^H\Sigma^{-1}\mathbf{h}'(\mathbf{x})d\mathbf{x}}{|d\mathbf{x}|} \\ \Rightarrow \lim_{|d\mathbf{x}| \to 0} \frac{\mathbf{h}(\mathbf{x})^T\Sigma^{-*}\mathbf{h}'(\mathbf{x})^*d\mathbf{x}^* + \mathbf{h}(\mathbf{x})^H\Sigma^{-1}\mathbf{h}'(\mathbf{x})d\mathbf{x}}{|d\mathbf{x}|} $$
The second line above is obtained by expanding the first line, and the third line is obtained by neglecting the last term in the numerator of the second line (since it is quadratic in $d\mathbf{x}$) and transposing the expression for the first term (which can be done since each term is a scalar).
It seems to me that, if $\mathbf{x} \in \mathbb{R}^M$, then the Jacobian becomes $\mathbf{h}(\mathbf{x})^T\Sigma^{-*}\mathbf{h}'(\mathbf{x})^* + \mathbf{h}(\mathbf{x})^H\Sigma^{-1}\mathbf{h}'(\mathbf{x})$.
But I don't understand the case for $\mathbf{x} \in \mathbb{C}^M$. How do you reconcile that $d\mathbf{x}$ and $d\mathbf{x}^*$ show up here?
( I believe $\mathbf{h}$ is analytic for my case. It contains expressions of the form $x_1e^{ix_2}$.)