Show that the linear functions Df(a) and Dg(b) are inverse to each other I am stuck with this problem, and I hope to get some hint to help me get through.
Let $A \subseteq \mathbb R^p$, $f:A \to \mathbb R^p$ . Let the function $g:f(A) \to \mathbb R^p $ be the inverse to $f$ in the sense that $ f \circ g(x)=x , g \circ f(y)=y $ for all $x \in A$ and $y \in f(A)$. 
If $f$ is differentiable at a point $a \in A$, and if $g$ is differentiable at $b=f(a)$, show that the linear functions $Df(a)$ and $Dg(b)$ are inverse to each other; that is, $Df(a) \circ Dg(b)$ and $Dg(b) \circ Df(a)$ are the identity on $\mathbb R^p.$
 A: Let $U, V, W$ be an open sets in $\mathbb{R}^n, \mathbb{R}^m, \mathbb{R}^k$ and $f: U \rightarrow V, g: V \rightarrow W$ differentiable functions.  
If we write $f(x) = (f_1(x), ... , f_m(x))$ and $g(y) = (g_1(y), ... , g_k(y))$, then we have $g \circ f(x) = (g_1 \circ f(x), ... , g_k \circ f(x))$.
The derivative of $f$ at a given point $p \in U$ is by definition a certain linear transformation $Df(p): \mathbb{R}^n \rightarrow \mathbb{R}^m$. The standard bases of $\mathbb{R}^n$ and $\mathbb{R}^m$ allow us to identify $Df(p)$ with the $m$ by $n$ matrix
$$A = \begin{pmatrix} \frac{\partial f_1}{\partial x_1}(p) & \cdots & \frac{\partial f_1}{\partial x_n}(p) \\ \vdots & & \vdots \\ \frac{\partial f_m}{\partial x_1}(p) & \cdots & \frac{\partial f_m}{\partial x_n}(p) \end{pmatrix}$$
In turn, the derivative of $Dg(f(p))$ of $g$ at $f(p)$ is a linear transformation $\mathbb{R}^m \rightarrow \mathbb{R}^k$ can be identified with the $k$ by $m$ matrix
$$B = \begin{pmatrix} \frac{\partial g_1}{\partial y_1}(f(p)) & \cdots & \frac{\partial g_1}{\partial y_m}(f(p)) \\ \vdots & & \vdots \\ \frac{\partial g_k}{\partial y_1}(f(p)) & \cdots & \frac{\partial g_k}{\partial y_m}(f(p)) \end{pmatrix}$$
Finally, the derivative of $g \circ f$ at $p$ can be identified with the $k$ by $n$ matrix whose $ij$th entry is $\frac{\partial(g_i \circ f)}{\partial x_j}(p)$.
The linear transformation $\mathbb{R}^n \rightarrow \mathbb{R}^k$ corresponding to the product $BA$ is the composition $Dg(f(p)) \circ Df(p)$.  From multivariable calculus, the partial derivative of $g_i \circ f: \mathbb{R}^n \rightarrow \mathbb{R}$ with respect to $x_j$ at $p$ is 
$$\sum\limits_{l=1}^n \frac{\partial(g_i)}{\partial y_l}(f(p)) \frac{\partial f_l}{\partial x_j}(p)$$
This shows that the derivative of a composition is the composition of the derivatives, which combined with the fact that the derivative of the identity map is the identity map, gives you your result.
