For $f(x)=x$, the half-derivative of $f$ is
$$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} x = 2 \sqrt{\frac{x}{\pi}} \;.$$
Is there some geometric interpretation of (Q1) this specific derivative, and, (Q2) of the half-derivative more generally? I have read that fractional derivatives are nonlocal,
but it seems strange to me that integral derivatives
can be described in terms of local
geometry only, while fractional derivatives cannot be so described.
This would suggest an odd discontinuity between, say, $d^{1}$ and $d^{1.01}$.
This seems especially at odds with the many applications of fractional derivatives,
which (superficially) suggests continuity should reign.
I'd appreciate someone clearing up my elementary confusions—Thanks in advance!
Addendum (5Jan14).
@AlexR. found this geometric interpretation of
the fractional integral in Richard Herrmann's book,
Fractional Calculus: An Introduction for Physicists,World Scientific, 2011:
   

A possible mechanical interpretation of the half-derivative can be given in terms of Abel's solution to a classical problem  from the calculus of variations (the tautochrone problem).

Let there be a heavy particle which is constrained to slide without friction along
the curve $y=y(t)$  in uniform gravity to its
lowest point. Then, given a function $T(y)$ that specifies the total
time of descent for a given starting height what is an equation of the
curve that yields this result?

The principle of conservation of energy implies that the distance $S=S(t)$ travelled by the particle along the curve from the initial height $y_0$ satisfies the equation
$$\left(\frac{dS}{dt}\right)^2=2g(y_0-y).$$
This is equivalent to the integral equation
$$T(y_0)=\frac{1}{\sqrt{2g}}\int_0^{y_0}\frac{1}{(y_0-y)^{1/2}}\frac{dS}{dy}dy.$$
The r.h.s. of the latter equation is nothing else but the Riemann–Liouville fractional integral of $f=\pi^{1/2}(2g)^{-1/2}dS/dy$, i.e.
$$D^{-\alpha}f(x)=\frac{1}{\Gamma(\alpha)}\int_0^x (x-y)^{\alpha-1}f(y)dy$$
of the order $\alpha=1/2$.
The solution to Abel's integral equation $dS/dy$ can be now interpreted (up to a constant factor) as the half-derivative of $T=T(y_0)$.

