Derivate as a division of differentials It's always said that $\frac{df(x)}{dx}$ is the derivate of a function with respect to x and it musn't be understood as a division of $df$ and $dx$. 
I've read that it is true that $\frac{df(x)}{dx}$ is the quotient of $df$ and $dx$. But this cannot be generalised for $\frac{d^2f(x)}{dx^2}$.
Could someone explain me why does this happen?
 A: If we want to keep thing clean and formal, then $\frac{df(x)}{dx}$ is very simply not a fraction. Any similarity with the notation of fractions is simply a notational similarity and nothing more. You could just as well denote the same thing by $f'(x)$, and nothing would change mathematically. This is more fully discussed in the question "Is $dy/dx$ not a ratio" alluded to in the comment. 
Of course, the story does not end there, because people chose this (slightly misleading) notation with good reason. If you allow yourself some degree of inexactness, or use non-standard analysis, you can make sense of heuristics such as $\frac{df(x)}{dx} = \frac{f(x+dx)-f(x)}{dx}$. If you just pretend that $dx$ is a variable, and pass to the limit $dx \to 0$, then you retrive the standard definition of derivative. In numerical methods, people seem to actually do this with finite $dx$, up to some accuracy.
There is nothing to stop you from doing the same with second derivative (except the same doubts as for the first derivative, of course):
$$ \frac{d^2f(x)}{dx^2} = \frac{d(f(x+dx) - f(x))}{dx^2} = \frac{f(x+2dx)-2f(x+dx)+f(x)}{dx^2}$$ 
For numerical purpses, this should work, at least in principle. If $f$ is twice differentiable, then in the limit $dx \to 0$ (again, treating the last term as a fraction) you will recover the usual second derivative. If you do non-standard analysis, and take $dx$ infinitesimal, then you recover the second derivative up to infinitesimal error. One difference is perhaps that you have to be careful with interpretation of $d^2 f$.

Later addition: A difference between first and second derivative is that for the first, you have an "if and only if" type of statement. Not only is it true that if $f$ is differentiable then $\frac{df(x)}{dx} = \lim_{dx \to 0} \frac{f(x+dx)-dx}{dx}$, but also $f$ is differentiablee at $x$ if the limit exists. This fails for the second derivative. For instance, if $f(x) = |x|$ then $\frac{f(x+2dx)-2f(x+dx)+f(x)}{dx^2} = 0$ if $dx$ is sufficiently small (depending on $x$), but $f$ is not even once differentiable at $x=0$. I believe it may also fail for once differentiable $f$, but I can't think of an example (translating the ordinary definition would lead to $\lim_{h_1 \to 0} \lim_{h_2 \to 0} \frac{f(x+h_1+h_2) -f(x+h_1)-f(x+h_2) + f(x)}{h_1h_2})$
A: The derivative $\frac{df}{dx}$ is the incremental ratio:
$$\frac{df}{dx}=\frac{f(x+\Delta x)-f(x)}{\Delta x}$$ when $\Delta x\to 0$
The quotient $\frac{d^2 f}{dx^2}$ is the incremental ratio of $\frac{df}{dx}$, that means:
$$\frac{d^2 f}{dx^2}=\frac{d}{dx}\frac{df}{dx}$$
so:
$$\frac{d^2f}{dx^2}=\frac{f(x+2\Delta x)-2f(x+\Delta x)+f(x)}{\Delta x^2}$$ when $\Delta x\to 0$ which still is a quotient or the incremental ratio
