What is $\frac{{\partial f(\left| z \right|)}}{{\partial x}} = $? Let $f(z)$ is polynomial and $z=x+iy$ , $x,y\in R$.
What is $\frac{{\partial f(\left| z \right|)}}{{\partial x}} = $?
 A: $$\begin{align} & \frac{{\partial f(\left| z \right|)}}{{\partial x}}
\\ & =\frac{{\partial f(\sqrt{x^2+y^2})}}{{\partial x}}
\\ & =\frac{\partial f(\sqrt{x^2+y^2})}{\partial (\sqrt{x^2+y^2})}\cdot \frac{\partial (\sqrt{x^2+y^2})}{\partial x}
\\ & =\frac{d\{f(\sqrt{x^2+y^2})\}}{d(\sqrt{x^2+y^2})}\cdot \frac{\partial (\sqrt{x^2+y^2})}{\partial x}
\\ & =\color{red}{f'(\sqrt{x^2+y^2})\cdot \frac{x}{\sqrt{x^2+y^2}}}\end{align}$$
Hope this helps.
A: Since
$$
\left|z\right|^2=z\,\bar{z}
\implies2\left|z\right|\mathrm{d}\!\left|z\right|=\bar{z}\,\mathrm{d}z+z\,\mathrm{d}\bar{z}
$$
we get
$$
\begin{align}
\mathrm{d}\!\left|z\right|
&=\mathrm{Re}\left(\frac{\bar{z}\,\mathrm{d}z}{\left|z\right|}\right)\\
&=\mathrm{Re}\left(\frac{(x-iy)\,\mathrm{d}(x+iy)}{\left|z\right|}\right)\\
&=\frac{x}{\left|z\right|}\mathrm{d}x+\frac{y}{\left|z\right|}\mathrm{d}y
\end{align}
$$
Therefore,
$$
\frac{\partial}{\partial x}\left|z\right|=\frac{x}{\left|z\right|}
$$
Thus, the chain rule says
$$
\bbox[5px,border:2px solid #C0A000]{\frac{\partial f\!\left(\left|z\right|\right)}{\partial x}=f'\!\left(\left|z\right|\right)\frac{x}{\left|z\right|}}
$$
A: Suppose $\;f(z)=\sum\limits_{k=0}^n a_kz^k\;$ , then
$$f(|z|)=\sum\limits_{k=0}^n a_k|z|^k=\sum_{k=0}^n a_k(x^2+y^2)^{k/2}\implies$$
$$\implies\frac{\partial}{\partial x}f(|z|)=\sum_{k=1}^n k\,x\,\,a_k(x^2+y^2)^{\frac k2-1}$$
