Note - my original definition below was wrong. I hope this replacement is better.
The usual approximation to $f'(x)$ with step size $h$ is $D_h(f, x) = \frac{f(x+h)-f(x)}{h} $. This has so many nice properties that I will not state any of them except that $D_h(f, x) = f'(x) + \Theta(h) $ as $h \to 0$.
A better approximation is $D^{=}_h(f, x) = \frac{f(x+h)-f(x-h)}{2h} $ which satisfies $D^{=}_h(f, x) = f'(x) + \Theta(h^2) $ as $h \to 0$.
These depend on the orientation of the coordinate system. In trying to come up with a rotation-independent derivative, it occurred to me that the arc length is the obvious choice for rotation independence. So here is my definition:
(These next two paragraphs have been completely rewritten.)
Let $(x(t), y(t))$ be a parameterization of $f$. The simplest one is $x(t) = t, y(t) = f(t)$, but any will work. For example, for a circle, two possibilities are $(t, \sqrt{1-t^2})$ and $(\cos(t), \sin(t))$ (for different ranges of $t$).
I define $D^{s}_h(f,t) = \frac{\Delta y}{\Delta x} = \frac{y(t+a(f, t, h))-y(t)}{x(t+a(f, t, h))-x(t)} $ where $a(f, t, h)$ is such that the arc length of $f$ from $t$ to $t+a(f, t, h)$ is $h$. In otherwords, $\int_t^{t+a(f, t, h)} \sqrt{x'^2(u)+y'^2(u)}du =h $.
My general question, which I an too lazy and tired at the moment to attempt to solve, is what are the properties of $D^{s}_h(f,x)$? Is there a nice expansion of $D^{s}_h(f,x)$ in powers of $h$ with the coefficients being functions of $f$ and its derivatives? Is this of any real use? Is this actually easy to solve, so it is one of those problems which, as Littlewood said, "there is less to this than meets the eye"?
Thank you for your consideration.