Let $V$ and $U$ be open sets in ${\bf{R}}^n.$ Let $f: U\to V$ be a diffeomorphism. Show that $Df(x)$ is bijective for all $x \in U$ and that $Df^{-1}(f(x)) = Df(x)^{-1}.$
The question seems a bit odd. It's a direct result from the Inverse Function Theorem right?
For the first part shouldn't $f$ be a bijection already for it to even be a diffeomorphism?
${\displaystyle f\colon M\rightarrow N}$ is called a diffeomorphism if it is a bijection and its inverse ${\displaystyle f^{-1}\colon N\rightarrow M}$ is differentiable as well.
The second part seems to be something I'm not sure on how to approach. We want to show that the derivative of the inverse of $f$ applied to $f(x)$ is the inverse of the derivative? Here $Df$ is an operator (matrix?) so $Df^{-1}$ would just be the derivative of the inverse, but I'm a bit confused on what does the notation $Df(x)^{-1}$ mean? Also from the problem statement we know that $Df$ is invertible since it's a diffeomorphism and $\det(Df)) \ne 0$ right?