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Let $f(z)$ is polynomial and $z=x+iy$ , $x,y\in R$.

What is $\frac{{\partial f(\left| z \right|)}}{{\partial x}} = $?

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$$\begin{align} & \frac{{\partial f(\left| z \right|)}}{{\partial x}} \\ & =\frac{{\partial f(\sqrt{x^2+y^2})}}{{\partial x}} \\ & =\frac{\partial f(\sqrt{x^2+y^2})}{\partial (\sqrt{x^2+y^2})}\cdot \frac{\partial (\sqrt{x^2+y^2})}{\partial x} \\ & =\frac{d\{f(\sqrt{x^2+y^2})\}}{d(\sqrt{x^2+y^2})}\cdot \frac{\partial (\sqrt{x^2+y^2})}{\partial x} \\ & =\color{red}{f'(\sqrt{x^2+y^2})\cdot \frac{x}{\sqrt{x^2+y^2}}}\end{align}$$

Hope this helps.

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  • $\begingroup$ @Undersky You're welcome. $\endgroup$ – SchrodingersCat Mar 19 '16 at 12:56
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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 $$ \bbox[5px,border:2px solid #C0A000]{\frac{\partial f\!\left(\left|z\right|\right)}{\partial x}=f'\!\left(\left|z\right|\right)\frac{x}{\left|z\right|}} $$

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Suppose $\;f(z)=\sum\limits_{k=0}^n a_kz^k\;$ , then

$$f(|z|)=\sum\limits_{k=0}^n a_k|z|^k=\sum_{k=0}^n a_k(x^2+y^2)^{k/2}\implies$$

$$\implies\frac{\partial}{\partial x}f(|z|)=\sum_{k=1}^n k\,x\,\,a_k(x^2+y^2)^{\frac k2-1}$$

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