Differentiable at a point and invertible implies inverse is differentiable? If $f:D\to\Bbb C$ is invertible and real (complex) differentiable at $c$ with $f'(c)\ne0$, it is easy to prove that if $f^{-1}$ is continuous at $f(c)$ and defined in a neighborhood of $f(c)$, then it is also real (complex) differentiable at $f(c)$ with derivative $1/f'(c)$:
$$\lim_{x\to f(c)}\frac{f^{-1}(x)-f^{-1}(f(c))}{x-f(c)}=\lim_{x\to f(c)}\frac{f^{-1}(x)-c}{f(f^{-1}(x))-f(c)}=\frac1{f'(c)},$$
where the latter limit goes to $1/f'(c)$ because $f^{-1}(x)\to c$ as $x\to f(c)$ by the assumption of continuity.
My question is: What additional assumptions are necessary to also deduce the extra assumptions on $f^{-1}$? In the real case, I know that it is sufficient that $f$ be continuous in a neighborhood of $c$, because then $f$ is monotone on an interval around $c$ and $f^{-1}$ is monotone and continuous in an interval around $f(c)$. In the complex case is this assumption sufficient?
 A: In general on a topological space, a function is invertible iff f is a bijection. A real valued function is a bijection iff it's monotone increasing or decreasing on a closed interval.  It's proven in real analysis that if f is a monotonic function defined on an interval I, then f is differentiable almost everywhere on I i.e. f is nondifferentiable at at most a countably infinite number of points in the interval. (Monotonicity is actually a weaker condition then monotone increasing/decreasing as it requires only the function is nonincreasing/nondecreasing on the interval. However, the function isn't a bijection and therefore, doesn't necessarily have an inverse, if this condition isn't met.)  Obviously, monotonicity isn't an option with complex valued functions since the complex plane isn't ordered. 
In fact, for real valued functions,the result you're looking for is this one and you'll see the conditions for a real valued map is considerably stronger then for analytic functions in the complex plane :

Does that answer your question? 
