Understanding higher dimensional derivatives I'm having trouble understanding higher dimensional derivatives.
Suppose $f: \Bbb R \to \Bbb R$.  We say $f$ is differentiable at $x = c$ if $\lim \limits_{x \to c} \dfrac{f(x) - f(c)}{x - c}$ exists.  If the limit exists, we define $f'(c)$ as the value limit.  Since this limit is a number, we can equivalently define the derivative of $f(x)$ at $c$ as the number $q$ such that $\lim \limits_{x \to c} \dfrac{f(x) - f(c) - q(x - c)}{x - c} = 0$.  If such a $q$ exists, we define $f'(c)$ to be this $q$.  We can think of $q$ as a "linear transformation" from $\Bbb R \to \Bbb R$.  Why is it useful to think of $q$ in this way?  What's the point of thinking about it as a linear transformation?  I think it probably has something to do with the fact that $f(c) + q(x - c)$ is the tangent line.
Now, we can define the derivative of $g: \Bbb R^{n} \to \Bbb R^{m}$ in the same way, that is, if there is some linear transformation $T_{g}$ such that $\lim \limits_{x \to c} \dfrac{||g(x) - g(c) - T_{g}(x - c) ||}{||x - c||} = 0$, then we say $T_{g}$ is the derivative of $g$.  Again, what's the point of saying this is a linear transformation? I already know it has to be an $m \times n$ matrix based on the context, but I don't see why we care that it is a linear transformation.  Does it have something to do with the tangent plane?
 A: If we start out with $f:\mathbb{R}\to\mathbb{R}$, then $f'(c)$ is an approximation of how $f$ changes in a small interval around $x=c$.  For example, let $f(x)=x^3$, and $c=2$.  Then $f'(2)=12$.  Notice that $f(2.01)=8.120601$.  Then the change from $f(2)$ to $f(2.01)$ is $0.120601$.  This is approximately $12(.01)$.  
For higher dimensions, the derivative needs to be a transformation between $\mathbb{R}^n$ and $\mathbb{R}^m$.  For example, take $f(x,y)=(x+y,y^2)$.  Then the derivative is
$$
\begin{pmatrix}
1 & 1 \\ 0 & 2y
\end{pmatrix}.
$$
At $(x,y)=(2,1)$ this is
$$
\begin{pmatrix}
1 & 1 \\ 0 & 2
\end{pmatrix}.
$$
Moving on, $f(2,1)=(3,1)$ and $f(2.01,1.01)=(3.02,1.0201)$.  The change between the function values is $(0.02,0.0201)$.  Notice that
$$
\begin{pmatrix}
1 & 1 \\ 0 & 2
\end{pmatrix}
\begin{pmatrix}
0.01 \\ 0.01
\end{pmatrix}
=
\begin{pmatrix}
0.02 \\ 0.02
\end{pmatrix}.
$$
Again, very close.  So, the derivative is the linear transformation that most closely fits the function.  Since linear transformations are much easier to study than functions in general, we may learn a lot about the function from its derivatives.
A: The meaning of the derivative is that it provides linear approximations of a function. Using the language of asymptotic analysis, we can say that, if $f$ has  a derivative at $a$, then:
$$f(x)=f(a)+f'(a)(x-a)+o(x-a)$$
where $o(x-a)$ — the ‘error term’ when replacing $f(x)$ with $f(a)+f'(a)(x-a)$ — is such that:
$$\lim_{x\to a}\frac{o(x-a)}{x-a}=0,$$
which means, roughly, that  the error term is negligible w.r.t. $\lvert x-a\rvert$.
Similarly, the second derivatice adds some information on this error term:
$$f(x)=f(a)+f'(a)(x-a)+f''(a)\frac{x^2}{2}+o((x-α)^2).$$
This is a quadratic approximation, known as Taylor's formula at order $2 $.
There are similar formulae for functions of several variables, and the derivative of the one-variable case has to be replaced with the differential, which is a linear form on $\mathbf R^n$, with coefficients equal to the partial  derivatives of $f$. The second derivative becomes a quadratic form on  $\mathbf R^n$, defined by the matrix of second order partial derivatives.
