TITLE: Definition of differentiation in context of abstract algebra.
QUESTION [7 upvotes]: In any regular calculus or real analysis course, we learn the definition of the derivative of a function $f(x)$ as $$f^\prime (x)=\lim\limits_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ However while studying abstract algebra we come to know that differentiation is just like any operation (like addition, multiplication etc.) but on functions. So I want to know that is there a way to define an algebraic structure with the underlying set as the set of all differentiable functions and the operation of differentiation.
And also if it's possible to define differentiation in such a manner, how to connect it with the analytical definition of differentiation.

REPLY [2 votes]: I once went to a seminar on non-commutative geometry, where the following was explained:
Consider a commutative ring $R$ with invertible element $h\in R$ (I gathered that in physical applications $h$ is often Planck's constant). We may take the following quotient of the (non-commutative) polynomial ring in two variables: $$A=R[x,y]/(xy-yx+hy).$$  Let $d\colon A\mapsto A$ be an $R$- linear map satisfying: \begin{eqnarray*}dx&=&y,\\
d(ab) &=& (da)b + a(db), \qquad{\rm for\,\,all\,\,}a,b\in A,\\
d\lambda&=&0,\qquad{\rm for\,\,all\,\,}\lambda\in R.
\end{eqnarray*}
One can prove: $$
d(x^n) = \frac{(x+h)^n - x^n}h dx \qquad[1].
$$
Extending linearly, for an arbitrary polynomial $f(x)$ over $R$ we get:
$$
 df(x) = \frac{f(x+h) - f(x)}hdx.
$$
I thought this was a nice result.  As far as I understand, the moral was that replacing commutativity (e.g. $[x,dx]=0$) with the identity $[x,dx]=-hdx$, replaces the usual "continuous" differential (where $h\to0$), with this "discrete" one.
Proof of $[1]$:
From the defining relation of $A$ we have $(x+h)y=yx$.  Thus $(x+h)^iy=yx^i$ for all natural numbers $i$.  The $n=1$ case of $[1]$ is clear.  We proceed by induction, by assuming the $n=k-1$ case of $[1]$ and deducing:
\begin{eqnarray*}
d(x^k)&=&d(xx^{k-1})\\&=&
yx^{k-1}+x\frac{(x+h)^{k-1}-x^{k-1}}hy\\&=&
yx^{k-1}+\frac{(x+h)^{k}-x^{k}}hy-(x+h)^{k-1}y\\&=&
\frac{(x+h)^{k}-x^{k}}hy.
\end{eqnarray*}