For $f:A\rightarrow\mathbb{C}$ with $A\subseteq\mathbb{C}$ let $\overline{f}:A\rightarrow\mathbb{C},x\mapsto\overline{f(x)}$ be the conjugate function.
Show for some $f$ differentiable at $x_0\in A\subseteq\mathbb{R}$ that $\overline{f}$ is differentiable at $x_0$ with $\overline{f'}(x)=\overline{f'(x_0)}$.
I'm having some trouble understanding this exercise. Now I do suspect that I need to show, technically, that
$$lim_{x\rightarrow x_0}\frac{\overline{f}(x)-\overline{f}(x_0)}{x-x_0} = lim_{x\rightarrow x_0}\frac{\overline{f(x)-f(x_0)}}{x-x_0}$$
exists. However I'm not even sure what $\overline{f}$ really is supposed to be or how to work with the equation above. The numerator has some complex conjugate, the denominator real numbers—how do I make use of that or in other words, what is $\overline{f(x)-f(x_0)}$ in the first place?
As stated the result is false. A simple counter-example: $A=\mathbb C,f(z)=z$. $f$ is differentiable at all points but its conjugate is not. Perhaps the question is about derivative as a function of two real variables, not about complex differentiation at all.