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The usual definition of the derivative of a a function $f:\mathbb{R}\to\mathbb{R}$ is $$\frac{\text{d}f}{\text{d}x}=\lim_{h \to0}\frac{f(x+h)-f(x)}{h}.$$ Could we define a derivative but instead of having a small additive step in $f(x+h)$ have a multiplicative step $f(hx)$.

I've tried playing around with the following as a definition for a derivative but it doesn't seem to match up with the standard definiton in any way. $$\frac{\text{d}f}{\text{d}x}=\lim_{h \to1}\frac{f(xh)-f(x)}{h-1}.$$

Can we defined a derivate in a similar manner to this and do we get any interesting results?

T. Stark
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    Provided $x \neq 0$ and $f$ is differentiable at $x$, we may substitute $x(h-1) = k$ to obtain $$\lim_{h\to1}\frac{f(xh) - f(x)}{h-1} = x \lim_{k\to0} \frac{f(x+k)-f(x)}{k} = xf'(x). $$ – Sangchul Lee May 10 '19 at 21:13
  • See $\alpha$-derivatives and the references here. – Ivo Terek May 10 '19 at 21:34
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    The limit you suggested exists if and only if either $x=0$ or $f$ is differentiable at $x$. If the former, it is $0$, if the latter, it is $xf'(x)$. –  May 10 '19 at 21:41
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    @IvoTerek that is something along the lines of what I was looking for. I have also found this https://en.wikipedia.org/wiki/Multiplicative_calculus – T. Stark May 10 '19 at 23:21

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There are a couple of interesting concepts to look at which have a similar flavour to this idea.

The first that comes to mind is the q-derivative or sometimes called the Jackson derivative. This looks very similar to the form suggested in the question, with the $q$-derivative given by $$\left(\frac{d}{dx}\right)_q f(x)=\frac{f(qx)-f(x)}{qx-x} .$$

Another thing that might be of interested to read about as suggested by Ivo Terek is $\alpha$-derivative (concept).

T. Stark
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