Theorem of Diffeomorphism. I'm starting out in geometry, I dont particularly understand the notation of $df_x$. 
what exactly is this linear map? a matrix? 
Can someone please draw an analogy between this and basic "highschool"  differentiation? 
ie polynomials? 

 A: The notation $df_x$ is intended to suggest an analogy with differential forms and the usual derivative in $\mathbb{R}^n$. To see this, first, calculate by definition what should be the differential of the "coordinate curves", i.e.,
the image under a parametrization of your manifold of the curves
$f_i: t\mapsto (u_1,u_2,\ldots, t, u_n)$ where all the $u_i$ are constant. 
the differential $df_i$ is just the derivative of the function that ignores all coordinates but the $i$th, i.e., $\partial/\partial x_i$. Now ask yourself, in terms of vector spaces, in $\mathbb{R}^n$, what is the dual of the projection to the $i$th coordinate? In the notation of differential forms, this would just be the form $dx_i$. But then you also have (by the definition of the differential for manifolds) $df_i=dx_i$.
Now, work out what the coefficients of the matrix $df$ for a general function should be. You will find they are all in terms of the $\partial/\partial x_i$. The notation $df$ stresses this. 
A: $\newcommand{\Reals}{\mathbf{R}}$Consider the case $n = 1$, i.e., real-valued functions of one variable. If $f$ is differentiable at a point $a$, then $f'(a) = df_{a}$ isn't "really" a number but a linear transformation from the tangent space $T_{a} \Reals$ to the tangent space $T_{f(a)} \Reals$, namely, multiplication by $f'(a)$. (The function calculus students blithely learn to call $f'$ is in fact a linear transformation-valued function, or a matrix-valued function if we fix a coordinate system.)
Letting $\Delta x = x - a$ and $\Delta y = y - f(a)$ denote coordinates for $T_{a} \Reals$ and $T_{f(a)} \Reals$, respectively, the linear transformation $df_{a}$ acts by
$$
\Delta y = df_{a}(\Delta x) = f'(a)\, \Delta x.
$$
This equation expresses the approximate linearity of $f$ at $a$. Calculus students encounter this as the equation of the tangent line to the graph $y = f(x)$ at $\bigl(a, f(a)\bigr)$:
$$
y - f(a) = f'(a) (x - a),\quad\text{or}\quad y = f(a) + f'(a)(x - a).
$$
If $f$ is smooth and invertible in some interval $J$, and if $f^{-1}$ is also smooth (i.e., $f$ is a diffeomorphism from $J$ to $f(J)$), then
$$
x = (f^{-1} \circ f)(x)\quad\text{for all $x$ in $J$.}
$$
Using the chain rule to differentiate at $a$,
$$
1 = (f^{-1})'\bigl(f(a)\bigr) \cdot f'(a),
$$
or
$$
(f^{-1})'\bigl(f(a)\bigr) = \frac{1}{f'(a)} = \bigl(f'(a)\bigr)^{-1}.
$$
In particular, $df_{a} = f'(a)$ is invertible.
A: Let $A$ be open in $\Bbb{R}^{n}$; let $f: A \to \Bbb{R}^{m}$; let $a \in A$. Then $f$ is said to be differentiable at $a$ iff there is some linear map $\Bbb{R}^{n} \to \Bbb{R}^{m}$, denoted by $df_{a}$, such that
$$
|x-a|^{-1}[ f(x) - f(a) - df_{a}(x-a)] \to 0 
$$
as $x \to a$;
in this case, the map $df_{a}$ is called the ("the" is due to the fact that $df_{a}$ exists only if $df_{a}$ is unique) derivative of $f$ at $a$.
The Jacobian matrix of $f$ at $a$, denoted by $Df(a)$, is defined as the matrix of $df_{a}$ with respect to the usual bases. 
Hence some authors do not distinguish $df_{a}$ from $Df(a)$; confusion can very unlikely arise in most cases.
If $n=m=1$, then $Df(a)$ is naturally identified with a point of $\Bbb{R}$, which is denoted by $f'(a)$ and is called the derivative of $f$ at $a$; this is what one usually sees in elementary univariate calculus.
The above is one of the ways that explain.
If $I: \Bbb{R}^{n} \to \Bbb{R}^{m}$ is identity and if $a \in \Bbb{R}^{n}$, then $I$ is linear; hence
$$
|x-a|^{-1}[I(x) - I(a) - I(x-a)] = |x-a|^{-1}\cdot 0 = 0
$$
as $x \to a$; this argument applies to every $a \in \Bbb{R}$, so we have $dI_{a} = I$ for all $a \in \Bbb{R}^{n}$.
