The complex version of the chain rule

I want to prove the following equality:

\begin{eqnarray}
\frac{\partial}{\partial z} (g \circ f) = (\frac{\partial g}{\partial z} \frac{\partial f}{\partial z}) + (\frac{\partial g}{\partial \bar{z}} \frac{\partial \bar{f}}{\partial z})
\end{eqnarray}

So I decide to do the following:

\begin{eqnarray}
\frac{\partial}{\partial z} (g \circ f) = \frac{1}{2}[(\frac{\partial g}{\partial x} \circ f)(\frac{\partial f}{\partial x}) + \frac{1}{i}(\frac{\partial g}{\partial y} \circ f)(\frac{\partial f}{\partial y})]
\end{eqnarray}

but the thing is that I am doing something wrong here since I don’t get any conjugate function and any derivative with respect to $\bar{z}$ so Can someone help me to see where I am wrong and fix it please?

In fact I don’t see what to do next, so I appreciate your help.

Thanks a lot in advance.

Edition:

What I’ve got so far is the following:

$$\frac{1}{2}[(\frac{\partial g}{\partial x} \circ f + \frac{\partial g}{\partial y} \circ f)\frac{\partial f}{\partial z} ]$$

but I’m still stuck.

Solutions Collecting From Web of "The complex version of the chain rule"

The question is taken in the context of Wirtinger Derivatives.

To that end, we let $g$ and $f$ be functions of both $z$ and $\bar z$. Then, the composite function $g\circ f$ can be expressed as

$$g\circ f=g(f(z,\bar z),\bar f(z,\bar z))$$

The partial derivative of $g\circ f$ with respect to $z$ is then given by

$$\begin{align}
\frac{\partial (g\circ f)}{\partial z}&=\frac{\partial (g(f(z,\bar z),\bar f(z,\bar z))}{\partial z}\\\\
&=\left.\frac{\partial g(w,\bar w)}{\partial w}\right|_{w=f(z,\bar z)}\times \frac{\partial f(z,\bar z)}{\partial z}+\left.\frac{\partial g(w,\bar w)}{\partial \bar w}\right|_{\bar w=\bar f(z,\bar z)}\times \frac{\partial \bar f(z,\bar z)}{\partial z}\\\\
&=\left(\frac{\partial g}{\partial z}\circ f\right)\frac{\partial f}{\partial z}+\left(\frac{\partial g}{\partial \bar z}\circ f\right)\frac{\partial \bar f}{\partial z}
\end{align}$$

You do some mistakes. Note that $$\frac{\partial}{\partial x} (g \circ f) = (\frac{\partial g}{\partial x} \circ f)(\frac{\partial f}{\partial x})+(\frac{\partial g}{\partial y} \circ f)(\frac{\partial f}{\partial x})$$ and $$\frac{\partial}{\partial y} (g \circ f) = (\frac{\partial g}{\partial x} \circ f)(\frac{\partial f}{\partial y})+(\frac{\partial g}{\partial y} \circ f)(\frac{\partial f}{\partial y}).$$ Now you should be able to finish your calculations.