derivative of a function composition which contain complex matrix

Question

I have a function $f(x)=2real(\mathbf{y}^H\mathbf{A(x)}\mathbf{h})-\mathbf{h}^H\mathbf{A(x)}^H\mathbf{A(x)}\mathbf{h}$, and I need to get the derivative in terms of $x$. I may resort to the chain rule, but here the matrix $\mathbf{A}$ is complex, and $x$, $f(x)$ are real. If I apply chain rule, I will get $\frac{\partial f(\mathbf{A(x))}}{\partial\mathbf{A}}=2real(\mathbf{y}\mathbf{h}^H)-\mathbf{h}\mathbf{h}^H\mathbf{A(x)}$ and $\frac{\partial\mathbf{A(x)}}{\partial\mathbf{x}}=\mathbf{B}$. And $\frac{\partial f(\mathbf{A(x))}}{\partial\mathbf{A}}\frac{\partial\mathbf{A(x)}}{\partial\mathbf{x}}=trace(2real(\mathbf{y}\mathbf{h}^H)\mathbf{h}\mathbf{h}^H\mathbf{A(x)}\mathbf{B})$ is absolutely not real. But since $f(\mathbf{A(x)})$ is real and $x$ is real, the derivative should be real. How can I get the derivative, $\frac{\partial f(\mathbf{A(x))}}{\partial\mathbf{x}}$?

Answer 1

You'll need the full complex chain rule àla $$\frac{\partial f(y(x))}{\partial x} = \frac{\partial f(y(x))}{\partial y} \frac{\partial y(x)}{\partial x} + \frac{\partial f(y(x))}{\partial y^*} \frac{\partial y^*(x)}{\partial x},$$ where $(\cdot)^*$ denotes complex conjugation. You've used only the first part, assuming that $y^*$ is independent of $x$.

As another hint, expanding $2 {\rm real}\{y\}$ into $2 {\rm real}\{y\} = y + y^*$ might help as well.