Functions of complex variable and its differentiation w=f(z)   (complex function)
w=u +iv (u, v are funtions of x and y)
How to find dw/dz in polar form?
Can someone please explain me the chain rule expansion involved in this
The derivative should be equal to partial differential of w in terms of r, theta and x
and z=re^(itheta)
 A: 
The total differential of $w=f(z)=:f(x,y)=u(x,y)+i(v,x)$ is
  \begin{align*}
df &= \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y}dy\\
&=\left(\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}\right)dx + \left(\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y}\right)dy\tag{1}
\end{align*}

With $x=r\cos \theta$ and $y=r\sin\theta$ we obtain
\begin{align*}
\frac{\partial u}{\partial x}
&=\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial \theta}\frac{\partial\theta}{\partial x}
=\frac{\partial u}{\partial r}\frac{x}{\sqrt{x^2+y^2}}-\frac{\partial u}{\partial \theta}\frac{y}{x^2+y^2}\\
&=\cos\theta\frac{\partial u}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial u}{\partial \theta}\tag{2}\\
\frac{\partial u}{\partial y}
&=\frac{\partial u}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial u}{\partial \theta}\frac{\partial\theta}{\partial y}
=\frac{\partial u}{\partial r}\frac{y}{\sqrt{x^2+y^2}}+\frac{\partial u}{\partial \theta}\frac{x}{x^2+y^2}\\
&=\sin\theta\frac{\partial u}{\partial r}+\frac{1}{r}\cos\theta\frac{\partial u}{\partial \theta}\tag{3}
\end{align*}
and similarly with $v(x,y)$ instead of $u(x,y)$.
We obtain with $x=x(r,\theta)=r\cos \theta$ and $y=y(r,\theta)=r\sin\theta$
\begin{align*}
dx&=\frac{\partial x}{\partial r} dr+\frac{\partial x}{\partial \theta}d\theta
=\cos\theta\,dr-r\sin\theta\,d\theta\tag{4}\\
dy&=\frac{\partial y}{\partial r} dr+\frac{\partial y}{\partial \theta}d\theta
=sin\theta\,dr+r\cos\theta\,d\theta\tag{5}\\
\end{align*}

Putting (2) - (5) into (1) we obtain
  \begin{align*}
\color{blue}{df}&\color{blue}{=\left(\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}\right)dx + \left(\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y}\right)dy}\\
&=\left(\cos\theta\frac{\partial u}{\partial r}-\frac{1}{r}\sin\theta \frac{\partial u}{\partial \theta}\right.\\
&\qquad\left.+i\left(\cos\theta\frac{\partial v}{\partial r}-\frac{1}{r}\sin \theta\frac{\partial v}{\partial \theta}\right)\right)\left(\cos\theta\,dr-r\sin\theta\,d\theta\right)\\
&\quad+\left(\sin\theta\frac{\partial u}{\partial r}+\frac{1}{r}\cos\theta \frac{\partial u}{\partial \theta}\right.\\
&\qquad\left.+i\left(\sin\theta\frac{\partial v}{\partial r}+\frac{1}{r}\cos \theta\frac{\partial v}{\partial \theta}\right)\right)\left(\sin\theta\,dr+r\cos\theta\,d\theta\right)\\
&=1\cdot\frac{\partial u}{\partial r}dr+0\cdot\frac{\partial u}{\partial \theta}dr
+0\cdot\frac{\partial u}{\partial r}d\theta+1\cdot\frac{\partial u}{\partial \theta}d\theta\\
&\quad+i\left(1\cdot\frac{\partial v}{\partial r}dr+0\cdot\frac{\partial v}{\partial \theta}dr
+0\cdot\frac{\partial v}{\partial r}d\theta+1\cdot\frac{\partial v}{\partial \theta}d\theta\right)\tag{6}\\
&\,\,\color{blue}{=\left(\frac{\partial u}{\partial r}+i\frac{\partial v}{\partial r}\right)dr+\left(\frac{\partial u}{\partial \theta}+i\frac{\partial v}{\partial \theta}\right)d\theta}
\end{align*}
  In (6) we use $\sin^2 \theta+\cos^2 \theta=1$ (factor $1$) and observe that terms cancel (factor $0$).

