The derivative as a linear transform I'm having trouble wrapping my head around thinking about the derivative as a linear transform.
Here is an example I came up with to try and understand it better.
Let $f: \mathbb{R}^2 \to \mathbb{R}^3$ be defined by $f(x,y) = (x+y,xy,x^2-y^2)$
Then the derivative of $f$ at $\vec{x_0} = \left[\matrix{
   x_0 \cr
   y_0 \cr
}\right]$ is given by: $f'_\vec{x_0} = \left[\begin{array}{*{20}{c}}
{1}&{1}\\
{y_0}&{x_0}\\
{2x_0}&{-2y_0}\\
\end{array}\right]
$ 
This means that $f'_\vec{x_0}$ is the linear transformation that takes a vector $\vec{v} \in \mathbb{R}^2$ and sends it to $f'_\vec{x_0}\vec{v} \in \mathbb{R}^3$. 
Here are my questions:
How do I interpret the derivative in higher dimensions? In one dimension we can think of the derivative as the slope of a tangent line. Is there a similar notion for higher dimensions? I'm looking at the above matrix and trying to attach some meaning to it, but I can't come up with anything.
What does  $f'_\vec{x_0}\vec{v}$ represent? Is it just the derivative of $f$ at $\vec{x_0}$ in the direction of $\vec{v}$? Do the components of this vector represent how fast $f$ is changing along each axis?
 A: The derivative (if exists) is the best local linear approximation in the following sense
$$f(x_0+h)=f(x_0)+f'(x_0)(h)+o(h),$$
or equivalently,
$$\lim_{h\to 0}\frac{f(x_0+h)-(f(x_0)+f'(x_0)(h))}{\|h\|}=0.$$
I.e. $f(x_0+h)-(f(x_0)+f'(x_0)(h))$ (function - approximation) is very small. So small that even divided by another small thing $\to 0$.
A: Are you familiar with the concept of a curve? A curve is a function $g:\mathbb R \rightarrow \mathbb R^3, t \mapsto g(t)$. As the parameter $t$ runs through $\mathbb R$, the points $g(t)$ run through the points of the curve in $\mathbb R^3$. With this concept, you can describe all kinds of curves - lines, circles, spirals, etc.
Using the visualization described above (as $t$ runs continuously from $-\infty$ to $+\infty$, $g(t)$ traces out the curve in 3-space), it's "intuitively clear" :-) that $g'(t_0)$ is the tangent vector at the curve in the point $g(t_0)$.
We can do the same thing with your map $f$. If $f$ is well behaved, then $M := f(\mathbb R^2) \subseteq \mathbb R^3$ is kind of a surface in space, like a sphere, or a piece thereof. If we fix an $x \in \mathbb R^2$ and a $v \in \mathbb R^2$, then the function $g:\mathbb R \rightarrow \mathbb R^3, t \mapsto f(x + tv)$ is a curve in 3-space, and by construction, this curve runs entirely inside $M$.
Now, using the chain rule, we find $g'(t) = f'(x+tv)v$, and in particular $g'(0) = f'(x)v$. So we see that $f'(x)v$ is the tangent of a curve on $M$ through $f(x) \in M$. All those curves form the tangent plane at $M$ in $f(x)$. More specifically, we can write this tangent plane as $P := f(x) + f'(x)\mathbb R^2 \subseteq \mathbb R^3$. Please convince yourself that the set $P$ is an affine plane in 3-space.
So, in some sense, $f'(x)$ is the slope of the tangent plane at $M$ through the point $f(x)$. With some more analytical machinery, it's possible to formalize this analogy to the one-dimensional case much better.
Another instructive example is $h:\mathbb R^2 \rightarrow \mathbb R^3, (u,v) \mapsto (u, v, \sin(u) + \cos(v))$.
