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.
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exp8j
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5You should read about derivations – Alex Jun 25 '19 at 18:31
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@Alex: Thank you, so it seems i have to dive into the area called differential algebra! – exp8j Jun 25 '19 at 19:23
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But in the meantime, is there a brief answer to my question? – exp8j Jun 25 '19 at 19:23
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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 '19 at 19:56
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I don't see why would it imply, for example, that $\exp'=\exp$. – Botond Jun 26 '19 at 06:59
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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.
Simon Segert
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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? – exp8j Jun 26 '19 at 17:28
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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. – Simon Segert Jun 26 '19 at 19:40