This is part of the proof of the inverse function theorem:
Let $U,V\subset R^n$ and $f\colon U\to V$ a differentiable bijection. Also assume that $f$ is $C^1$ and $f^{-1}$ is differentiable. Show that $D(f^{-1})$ is continuous.
Using the chain rule, I have derived that $Df^{-1}(f(x))=Df(x)^{-1}$. Since $f$ is $C^1$, we know that $Df(x)$ is continuous. So what I basically have to show is that if a function $g\colon U\to V$ is continuous (with $U,V$ satisfying the above conditions), then its inverse $g^{-1}$ is continuous too. I have seen someone mention the invariance of domain theorem on this site, but I feel like that is too advanced for me now, in the sense that our teacher actually meant something easier (since this is a part of a small homework exercise). Unless I'm wrong, there should be a relatively easy to show this. Any hint?
Just remark that $Df^{-1}(x)=Df(f^{-1}(x))^{-1}$, $x\rightarrow f^{-1}(x), x\rightarrow Df(x)$ and $A\rightarrow A^{-1}$ defined on $Gl(\mathbb{R}^n)$ are continuous and $x\rightarrow Df^{-1}(x)$ is the composition of these functions.