TITLE: Hyper Derivative definition.
QUESTION [19 upvotes]: Following some thoughts in real analysis and in analogy with the classical definition of the derivative
$$
f^\prime(x) = \lim_{h\rightarrow 0}{f(x+h)-f(x)\over h}
$$
I considered the following hyper-derivative definition:
$$
f^\nabla(x) = \lim_{h\rightarrow 1}\log_h{f(x\cdot h)\over f(x)}
$$
It is easily proved then that
$$
(f\cdot g)^\nabla = f^\nabla + g^\nabla
$$
I would like please if someone can give some expert opinion in the following:

Is the hyper derivative definition well defined? or somedody can see a potential problem?
Can someone provide a geometric interpretation for the hyper derivative definition?

REPLY [30 votes]: What is $\boldsymbol{\,f^\nabla(x)}$?
$$
\begin{align}
f^\nabla(x)
&=\lim_{h\to1}\log_h\left(\frac{f(xh)}{f(x)}\right)\\
&=\lim_{h\to1}\frac{\log(f(xh))-\log(f(x))}{\log(h)}\\
&=\lim_{\delta\to0}\frac{\log(f(x+\delta\,x))-\log(f(x))}{\log(1+\delta)}\\
&=\lim_{\delta\to0}\frac{\delta\,x}{\log(1+\delta)}\,\lim_{\delta\to0}\frac{\log(f(x+\delta\,x))-\log(f(x))}{\delta \,x}\\
&=x\,\frac{\mathrm{d}}{\mathrm{d}x}\log(f(x))\\
&=x\,\frac{f'(x)}{f(x)}
\end{align}
$$
which is well-defined as long as $\,f'(x)$ exists and $f(x)\ne0$.
Furthermore, if $f'(0)$ exists and $f(0)=0$, then $f^\nabla(0)=1$.

Geometric Interpretation
Extending the result above,
$$
\begin{align}
f^\nabla(x)
&=x\,\frac{\mathrm{d}}{\mathrm{d}x}\log(f(x))\\
&=\frac{\mathrm{d}\log(f(x))}{\mathrm{d}\log(x)}
\end{align}
$$
which would be the slope of the graph of $f(x)$ in a log-log plot.