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Given a real number $x>0$, define it's "derivative" as follows:

$$x' = x \log(x).$$

We can easily show that $$(xy)' = x'y+xy'$$ for all $x,y > 0$. In particular: $$(xy)' = xy \log(xy) = xy(\log(x)+\log(y)) = xy \log(x)+xy\log(y) = x'y+xy'$$

So, we started with something, namely logarithms, that turns multiplication into addition, and we obtained something, namely "derivatives" in the above sense, satisfying the Leibniz law, and these are related by $x'/x = \log(x)$.

This suggests that we try to go the other way. Given a function $f$ on the real line, define it's "loogarithm" to be $$\mathrm{loog}(f) = \frac{f'}{f},$$ where $f'$ is the ordinary derivative. If we're willing to assume $f$ is positive, then the loogarithm can be understood as the act of taking the ordinary logarithm and then differentiating. In more detail: $$\mathrm{loog}(f) = (\log(f))'.$$ So basically, $\mathrm{loog}$ is the well-known technique of logarithmic differentiation.

It's straightforward to check that $$\mathrm{loog}(fg) = \mathrm{loog}(f)+\mathrm{loog}(g),$$ which is what makes that technique work, after all.

The above observations generalize readily to derivations on commutative rings/fields.

In particular we're given a function $\log : R \rightarrow R$ on a commutative ring satisfying $\log(xy) = \log(x)+\log(y)$ and $\log(1) = 0$, then we can get a notion of differentiation on $R$ via $x' = x \log(x)$, and it will always satisfy Leibniz. Conversely, the formula $\log(x) = x'/x$ allows us to turn anything satisfying Leibniz into a notion of logarithm that's defined for all units.

Question. Is there any theory surrounding this ability to go back and forth between "logarithms" and "differentiation" in this way? It seems to be rather important.

What follows is a long discussion about this kind of thing.

We might try inverting $\mathrm{loog}$ to get some kind of exponentiation function. Unfortunately, unlike $\log$, the loogarithm is not invertible because differentiation does funny things. Oh well, let's just add the basepoint from which we're going to start integrating as an extra parameter $p$: $$a_p^f := a^{\int_p f}.$$

This exponentiation operator occurs in the method of integrating factors. Some of it's properties are familiar looking:

$$a_p^0 = 1, \qquad a_p^{f+g} = a_p^f a_p^g, \qquad a_p^{-f} = \frac{1}{a_p^f}$$

Others look a bit strange, for example:

$$a^1_p(x) = a^{x-p}$$

And I haven't been able to find anything worthwhile to say about $a_p^{fg}$.

The method of integrating factors works because of the statement: $$(e_p^f)' = f \cdot e_p^f.$$

That means that we can transform the DE $$f' + pf = q$$ into $$f'e_0^p + pfe_0^p = qe_0^p$$ which becomes $$(f e_0^p)' = qe_0^p.$$ So $$fe_0^p = f(0)+\int_0 qe_0^p$$ which yields $$f = e_0^{-p}\left(f(0)+\int_0 qe_0^p\right)$$

Going back to the idea that numbers can be "differentiated" via $x' = x \log(x)$, the above technique can be applied to solve equations of the form $x \log(x)+px = q.$ In particular, multiply both sides by $e^p$. We obtain:

$$x' e^p + pxe^p = qe^p$$

So $(xe^p)' = qe^p.$

To finish the problem, we can use the Lambert $W$ function and this, or just define our function and call it "integration."

This post is getting pretty long so I might just leave it there.

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    $\begingroup$ I don't think there is much theory. The problem with those definitions is that $(x+y)'\eq x'+y'$. The arithmetic derivative has the same problem. $\endgroup$ – Professor Vector Sep 16 '17 at 17:07
  • $\begingroup$ As Professor Vector already remarked, you have that $(x+y)'\neq x'+y'$, which is probably what causes that you cannot find something useful for $a^{fg}_p$. $\endgroup$ – Sergio Enrique Yarza Acuña Sep 16 '17 at 17:32
  • $\begingroup$ @Chappers It seems you forgot to put your question before the question mark. What was it you wanted to know? $\endgroup$ – Professor Vector Sep 16 '17 at 19:33

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