TITLE: Partial derivatives in $\mathbb{C}^n$
QUESTION [5 upvotes]: I'm trying to figure out an equality from a proof by Griffiths and Harris to the holomorphic inverse function theorem (in Principles of Algebraic Geometry). They state:
$$\frac{\partial}{\partial \overline{z}_i}(f^{-1}(f(z))) = \sum_k\frac{\partial f_j^{-1}}{\partial z_k} \cdot \frac{\partial f_k}{\partial \overline{z}_i} + \sum_k\frac{\partial f_j^{-1}}{\partial \overline{z}_k} \cdot \frac{\partial \overline{f}_k}{\partial \overline{z}_i}$$
I'm a bit confused about how this deriviation came about... I would expect from the chain rule to have something in the lines of $ \frac{\partial f_j^{-1}}{\partial f_k} \cdot \frac{\partial f_k}{\partial \overline{z}_i}$. Or do i have it all wrong?
And how did this sum get here? I understand the definition
$$df = \sum_k\frac{\partial f}{\partial z_k}dz_k + \sum_k\frac{\partial f}{\partial \overline{z}_k}d\overline{z}_k $$
but how does this relate to partial derivatives such as the one above?
Thank you.

REPLY [3 votes]: The chain rule is $$\frac{\partial (f\circ g)}{\partial z}=\left(\frac{\partial f}{\partial z}\circ g\right)\frac{\partial g}{\partial z}+\left(\frac{\partial f}{\partial \overline{z}}\circ g\right)\frac{\partial \overline{g}}{\partial z}.$$
This essentially comes from seeing $f$ as a function in the variables $z$ and $\overline{z}$, and so $f\circ g(z)=f(g(z,\overline{z}),\overline{g(z,\overline{z})}).$ See http://en.wikipedia.org/wiki/Wirtinger_derivatives, for example.
Note: This is for functions of one variable. The general case like you have is very similar.