For each $a \in U$ we have $Df(a):\mathbb R^n \to \mathbb R^n$ is a linear isomorphism An invertible function $f: U \subseteq \mathbb R^n \to V \subseteq \mathbb R^n$ is given and we know that $f$ and $f^{-1}$ both are differentiable.  
Prove that for each $a \in U$ we have $Df(a):\mathbb R^n \to \mathbb R^n$ is a linear isomorphism. Then for $b \in V$, find $Df^{-1}(b)$.  
I can't even understand how this is possible! For example, How can derivatives of $f:\mathbb R \to \mathbb R$ be isomorphic to $\mathbb R$ itself?
 A: A function $f: U \subset \mathbb{R}^n \to V \subset \mathbb{R}^m$ is said to be differentiable if there exists a linear map $\lambda$ such that,
$$ \lim_{x \to p} \frac{\|f(x) - f(p) - \lambda(x-p)\|}{\|x-p\|} = 0 \ \ \ \ \textbf{or} \ \ \ \ \lim_{h \to 0} \frac{\|f(p+h) - f(p) -\lambda(h) \|}{\|h\|} = 0$$
for all $p \in U$. What this means is that $\lambda$ is a linear map which approximates the change in $f$ for points $x$ sufficiently close to $p$. By the above definition, $\lambda$ is also unique. One can show that,
$$\tilde{\lambda}(\textbf{h}) =  \begin{pmatrix} \frac{\partial f^1}{\partial x^1} & \cdots & \frac{\partial f^1}{\partial x^n} \\ \vdots \\ \frac{\partial f^m}{\partial x^1} & \cdots & \frac{\partial f^m}{\partial x^n} \end{pmatrix} \cdot \textbf{h}, \ \ \ \textrm{where} \ \ \ \ f = \begin{pmatrix} f^1 \\ f^2 \\ \vdots \\ f^m \end{pmatrix}$$
is a map which satisfies the above defintion and so $\tilde{\lambda} = \lambda$. From the definition above it is clear that $\lambda$ is linear. Instead of using $\lambda$ we adopt a different notation $D_pf$ or $Df(p)$. Typically one makes no distinction between a linear map (on a f.d.v.s) but since this may be your first time seeing this, one can define the matrix above as the Jacobian $\textbf{J}(Df(p))$ and let $Df(p)$ be the corresponding linear operator. 
\begin{align*} Df(p): &\mathbb{R}^n \to \mathbb{R}^m \\& v \mapsto \textbf{J}_p(v) \end{align*}
By the chain rule, given $\gamma: (- \epsilon, \epsilon) \to U$ is a smooth curve with $\gamma(0) = p$ and $\gamma'(0) = v$ we have a relation,
\begin{align*} \frac{d}{dt}\Bigr|_{t = 0} (f \circ \gamma)(t) &= Df(\gamma(0)) \cdot \gamma'(0) \\ & = Df(p) \cdot v \\ & = \textbf{J}_p(v) \end{align*}
A: You have $f \circ f^{-1} = I$ and $f^{-1}\circ f = I$. Take the derivative of both sides in each equation and apply the chain rule. Conclude.
