1
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ You should read about derivations $\endgroup$ – Alex Jun 25 at 18:31
  • $\begingroup$ @Alex: Thank you, so it seems i have to dive into the area called differential algebra! $\endgroup$ – exp8j Jun 25 at 19:23
  • $\begingroup$ But in the meantime, is there a brief answer to my question? $\endgroup$ – exp8j Jun 25 at 19:23
  • $\begingroup$ 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. $\endgroup$ – Bill Dubuque Jun 25 at 19:56
  • $\begingroup$ I don't see why would it imply, for example, that $\exp'=\exp$. $\endgroup$ – Botond Jun 26 at 6:59
2
$\begingroup$

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.

Also, note that this is not a definition of the derivative, since simply writing down a list of properties does not guarantee that there actually exists an operator with those properties.

$\endgroup$
  • $\begingroup$ I do not understand your last sentence. If it is proved that the usual derivative, and only this, is the operator that satisfies the properties, then why not consider them as an equivalent definition of the derivative? $\endgroup$ – exp8j Jun 26 at 17:28
  • $\begingroup$ This might merely be a semantic quibble. What I mean is that if you try to define the derivative as the unique operator satisfying your three properties, then showing that this definition is well-formed (i.e. uniquely specifies D) requires that the derivative has already been defined in the usual way. For example, in a textbook, it would be difficult to give this definition before giving the usual definition of a derivative. $\endgroup$ – Mike Hawk Jun 26 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.