Is there any meaning to this "Super Derivative" operation I invented?

I've thought about this for a few days now, I didn't originally intend to answer my own question but it seems best to write this as an answer rather than add to the question. I think there is nice interpretation in the following: $$ f(x) = \lim_{h \to 0} \frac{e^{h f(x)}-1}{h} $$ also consider the Abel shift operator $$ e^{h D_x}f(x) = f(x+h) $$ from the limit form of the derivative we have (in the sense of an operator) $$ D_x = \lim_{h \to 0} \frac{e^{h D_x}-e^{0 D_x}}{h} = \lim_{h \to 0} \frac{e^{h D_x}-1}{h} $$ now we can also manipulate the first equation to get $$ \log f(x) = \lim_{h \to 0} \frac{f^h(x)-1}{h} $$ so by (a very fuzzy) extrapolation, we might have $$ \log(D_x) = \lim_{h \to 0} \frac{D_x^h-1}{h} $$ and applying that to a function we now get $$ \log(D_x) f(x) = \lim_{h \to 0} \frac{D_x^h f(x)-f(x)}{h} $$ which is the $\alpha = 0$ case of the 'super-derivative'. So one interpretation of this case is the logarithm of the derivative? If we apply the log-derivative to a fractional derivative then we have $$ \log(D_x) D^\alpha_x f(x) = \lim_{h \to 0} \frac{D_x^h D^\alpha_x f(x)-D^\alpha_x f(x)}{h} $$ there might be a question of the validity of $D_x^h D^\alpha_x = D_x^{\alpha+h}$ which I believe isn't always true for fractional derivatives.

This interpretation would explain the $\log(x)$ type terms arising in the series above. I'd be interested to see if anyone has any comments on this? I'd love to see other similar interpretations or developments on this. What are the eigenfunctions for the $\log D_x$ operator for example? Can we form meaningful differential equations?

Edit: For some functions I have tried we do have the expected property $$ n \log(D_x) f(x) = \log(D_x^n) f(x) $$ with $$ \log(D_x^n) f(x) = \lim_{h \to 0} \frac{D_x^{n h} f(x)-f(x)}{h} $$

Seems like you have happened upon some relations similar to ones I've written about over several years. Try for starters the MSE-Q&A "Lie group heuristics for a raising operator for $(-1)^n \frac{d^n}{d\beta^n}\frac{x^\beta}{\beta!}|_{\beta=0}$." There are several posts on my blog (see my user page) on this topic, logarithm of the derivative operator (see also A238363 and links therein, a new one will be added soon, my latest blog post), and fractional differ-integral calculus.