Prove that $f(A)$ is an open set and $f^{-1}:f(A)\to A$ is differentiable. Let $A\subset \mathbb{R}^n$ an open set, and $f:A\to \mathbb{R}^n$ a one to one and continuously differentiable function so that $\det f'(x)\ne 0$ for all $x\in A$ Prove that $f(A)$ is an open set and $f^{-1}:f(A)\to A$ is differentiable.
Any idea on how to prove this? I´m totally lost.  (This seems related to the inverse function theorem). 
Any help or hint will be appreciated, thanks.
 A: The inverse function theorem tells you that for each $x\in A$, there is an open neighborhood $U$ of $x$ and an open neighborhood $W$ of $f(x)$ contained in $f(U)$ such that $f^{-1}:W\to U$ is differentiable.  So $f(A)$, being the union of these open neighborhoods $W$ around each point $f(x)\in f(A)$, is open.  Since differentiability is a local property, it also follows that $f^{-1}$ is differentiable as a function from all of $f(A)$ to $A$.

The original version of the question left out the assumption that the derivative of $f$ is continuous.  In fact, the result is still true without that hypothesis, and can be proven using more advanced techniques as follows.  The invariance of domain theorem (typically proved using methods from algebraic topology) states the following:

Suppose $A\subseteq\mathbb{R}^n$ is open and $f:A\to\mathbb{R}^n$ is a continuous injective map.  Then the image $f(A)$ is an open subset of $\mathbb{R}$, and the inverse map $f^{-1}:f(A)\to A$ is continuous.

In particular, this theorem applies to your function $f$, giving you that $f(A)$ is open and $f^{-1}$ is continuous.  To get that $f^{-1}$ is differentiable (and its derivative is the inverse of the derivative of $f$), you can use the same argument as in the proof of the inverse function theorem.
In fact, the result is true even if you drop the assumption that $f$ is injective, as long as you only ask for the conclusion to hold locally (i.e., every $x\in A$ has an open neighborhood $U$ such that $f(U)$ is open, $f|_U:U\to f(U)$ is injective, and $f|_U^{-1}:f(U)\to U$ is differentiable).  This result is even harder, and references to proofs are given in the answers to this question on MathOverflow.
