A question on generalization of the concept of derivative I am looking for some material to understand the process of generalization of the concept of derivative. I would not like to just read and apply the definition of the concept of differentiation in order to comprehend this generalization. I would like to work with Differential Calculus fluently so please forgive me if this is not a high-level question.
What I've read is that for functions $f:\mathbb R\to\mathbb R$ it would be equivalent the fact that the limit
$$
\lim_{h\to 0} \frac{f(x_0+h)-f(x_0)}{h}
$$
exists and the existence of a (unique) linear map $\lambda:\mathbb R\to\mathbb R$ (depending on $x_0$) such that
$$
\lim_{h\to 0} \frac{f(x_0+h)-f(x_0)-\lambda(h)}{h} = 0.
$$
Correct me if I'm wrong but only in the case $f:\mathbb R^1 \to \mathbb R^1$ and because of the definition of the function $f'$, it would be $\lambda(h):=h f'(x_0)$. But we use the equivalence with the next equation in order to generalize the concept for $\mathbb R^n\to \mathbb R$ (and also $\mathbb R^n\to\mathbb R^m$) functions:
$$
\lim_{h\to 0} \frac{f(x_0+h)-f(x_0)-\lambda(h)}{\lVert h\rVert} = 0\;\: (\in\mathbb R),
$$
which does not depend on the undefined operation of division $\frac{1}{h}$ in $\mathbb R^n$ (why then not requiring the existence of the limit $\lim_{h\to 0}(f(x_0+h)-f(x_0))/{\lVert h\rVert}$?).
Is this correct? Have I committed a lot of inaccuracies? Concluding, what I would like to mean by this question is that I want to understand every step taken in the method of generalization of the concept of derivative. Thanks in advance.
 A: Let $f:\mathbb R\to\mathbb R$ and $x$ in $\mathbb R$. The function $f$ is differentiable at $x$ if and only if there exists a linear function $g_x:\mathbb R\to\mathbb R$ and a function $\varepsilon_x:\mathbb R\to\mathbb R$ such that $\lim\limits_{h\to0}\varepsilon_x(h)=0$ and, for every $h$ in $\mathbb R$, 
$$
f(x+h)=f(x)+g_x(h)+|h|\varepsilon_x(h).
$$
When this happens, $g_x(h)=\lambda\,h$ for some real number $\lambda$, denoted $f'(x)$ and called the derivative of $f$ at $x$.
Let $f:\mathbb R^{\color{red}{n}}\to\mathbb R^m$ and $x$ in $\mathbb R^{\color{red}{n}}$. The function $f$ is differentiable at $x$ if and only if there exists a linear function $g_x:\mathbb R^{\color{red}{n}}\to\mathbb R^m$ and a function $\varepsilon_x:\mathbb R^{\color{red}{n}}\to\mathbb R^m$ such that $\lim\limits_{h\to0}\varepsilon_x(h)=0$ and, for every $h$ in $\mathbb R^{\color{red}{n}}$, 
$$
f(x+h)=f(x)+g_x(h)+|h|\varepsilon_x(h).
$$
When this happens, the linear function $g_x$ is  called the differential of $f$ at $x$. As every linear function from $\mathbb R^{\color{red}{n}}$ to $\mathbb R^m$, $g_x$ can be represented by a matrix $D_x$ of size $m\times {\color{red}{n}}$ such that, for every $h$ in $\mathbb R^{\color{red}{n}}$,
$$
g_x(h)=D_x\cdot h,
$$
and the coefficients of $D_x$ are more commonly denoted
$$
(D_x)_{k,\ell}=\frac{\partial f_k}{\partial x_\ell}(x).
$$
A: Here is a set of lecture notes that explains the generalisation of the derivative from functions mapping R to R, as you had asked for, to higher finite dimensions:
http://www.math.ufl.edu/~groisser/classes/handouts/differentiable.pdf
On your question about "how and why", there is no hard and fast rule about generalising derivatives, rather, we might ask what we want our new definition to satisfy. That document develops the idea of a 'good linear approximation'. First and foremost, to qualify as a valid generalisation it must include the case one has generalised from, and it must be a sensible definition (it doesn't contradict itself, for example). After these things are sorted, we might then ask for certain properties, and this will shape our definition.
For example, you might know that a differentiable function on the real line is also continuous. By requiring that our candidate derivative is a linear transformation, we get this pretty much for free in higher finite dimensions too. Note, that if we generalise our concept of the derivative to an infinite dimensional function space (quite often to the space of continously differentiable functions on a closed interval, $C^1[a,b]$), we must ask for more of our derivative (namely that it is a continuous linear transformation) to guarantee that the function(al) we are differentiating is also continuous. This is known as Fréchet differentiability, and is where alternative definitions that could fail this continuity requirement begin to emerge. Moreover, the derivative you are aware of is in the class of "Linear Differential Operators", which have a nice linear algebraic structure that we can make use of. You might have solved systems of differential equations in your linear algebra class.
We don't HAVE to define our derivative to be linear, but it is often useful to us to do so. We don't want to define something so weak that when we say a function is differentiable, it doesn't really help us say anything else about it. Wikipedia has an example here in its article on differential operators of a non-linear operator:
http://en.wikipedia.org/wiki/Differential_operator
If you study physics you'll certainly see a few of these. 
Does this help at all?
