# Can differentiation be defined in an algebraic way?

Is it possible to define the operation D of differentiation of real functions in an abstract way, as for example by the fundamental properties of the derivative:
D(f+g) = D(f) + D(g)
D(fg) = fD(g) + gD(f)
if f(x) = x, D(f) = 1 (to avoid the trivial D(f) = 0).

So if the above conditions hold, and assuming we are in the set of real differentiable functions, does it follow that D(f) is the known derivative of f?
Possibly more assumptions are required, so i'm asking for guidance.

It would be amazing if one could define differentiation in such an algebraic way.
Also possibly do the same for other operations, like integration and exponentiation.

• You should read about derivations – Alex Jun 25 at 18:31
• @Alex: Thank you, so it seems i have to dive into the area called differential algebra! – exp8j Jun 25 at 19:23
• But in the meantime, is there a brief answer to my question? – exp8j Jun 25 at 19:23
• You might find enlightening some of my posts where I mention / use purely algebraic derivatives, e.g. see here and here and here and here. – Bill Dubuque Jun 25 at 19:56
• I don't see why would it imply, for example, that $\exp'=\exp$. – Botond Jun 26 at 6:59

Yes it is a standard argument in differential geometry that these rules uniquely specify the derivative. Indeed, fix a differentiable function $$f$$ and a point $$p$$. By Taylor's theorem, we have $$f(x)=f(p)+(x-p)f'(p)+(x-p)h(x)$$ where $$h(p)=0$$. Then by applying the axioms in your question, it is simple to see that $$(Df)(p)=f'(p)$$ as desired.