Is the order of a derivative not additive under the Riemann-Liouville definition? I was doing some mathematics doodling today and I wrote down
$$\frac{d^{n}}{dx^{n}}x^n = n!.$$
Looking at this made me wonder if when $n = 1/2$, if $D^{1/2}_0\sqrt{x} = \Gamma(1/2)$. Where $D^{1/2}_{0}$ is the Riemann-Liouville derivative with order 1/2 and lower terminal $0$. Indeed, it appears that
$$D^{1/2}_0 \sqrt{x} = \frac{1}{\Gamma(3/2)} \frac{d}{dx}\int_{0}^{x}\frac{\sqrt{t}}{\sqrt{x-t}}dt= \frac{2}{\Gamma(1/2)} \frac{d}{dx} \frac{x\pi}{2} = \sqrt{\pi}$$
as expected. The Wikipedia entry (which I've verified) suggests that "$D^{1/2}_{0}D^{1/2}_{0} = \frac{d}{dx}$", that is, two half derivatives is equivalent to the usual derivative and verifies it in the case of $f(x) = x$ by computing $D_0^{1/2}D_0^{1/2}x = 1$.
However, for a constant $c$,
$$D_0^{1/2}c = \frac{1}{\Gamma(3/2)} \frac{d}{dx} \int_{0}^{x} \frac{c}{\sqrt{x-t}}dt = \frac{1}{\Gamma(3/2)}\frac{d}{dx} 2c\sqrt{x} = \frac{c}{\sqrt{\pi x}}.$$
Now using these facts, I was curious what function satisfied $D_{0}^{1/2}f(x) = 1$ and noticed that
$$D_{0}^{1/2}D_{0}^{1/2}\sqrt{x} = D_{0}^{1/2} \Gamma(1/2) = D_{0}^{1/2}\sqrt{\pi} = \frac{\sqrt{\pi}}{\sqrt{\pi x}} = \frac{1}{\sqrt{x}} \neq \frac{d}{dx}\sqrt{x} =\frac{1}{2\sqrt{x}}.$$
At this point I am confused and I would like to better understand what is going on.
Wikipedia aside, I thought I was 100% certain I remember reading that the primary advantage of using the Riemann-Liouville definition was that the composition of fractional derivative operators "added up" in the way you would like, but this seems to contradict this. If it doesn't, then what was the purpose of Wikipedia's repeated half differentiation of $x$?
I think I must have made a simple arithmetic error, but I cannot find where the missing factor of 2 would come from to fix this.
 A: Lets keep track of some facts-


*

*$\Gamma(3/2) = \frac{\sqrtπ}{2}$ and $\Gamma(1/2) = \sqrt{π}$. 

*$\frac d{dx} \sqrt{x}  = \frac1{2\sqrt x}$ (These two you got right.)

*$ D^{\alpha} f:= \frac d{dx}J^{1-\alpha}f = \frac{1}{\Gamma(1-\alpha)}\frac d{dx}\int_0^x \frac {f(t)}{(t-x)^{\alpha}} \, dt$ (here you have the wrong Gamma constant for $\alpha = 1/2$., e.g. see Wikipedia.)


Forgive me for dropping the 0 subscript, but here we see where the missing factor of 2 is. With this we see (where you can verify the integrals via e.g. Wolfram Alpha)
$$ D^{1/2}\sqrt x = \frac{1}{\Gamma(1/2)}\frac d{dx} \frac{xπ}2 = \frac{\sqrt{π}}{2}$$
which by the way is on the Wikipedia page, and similarly  (here you got the right answer somehow)
$$ D^{1/2} 1 = \frac{1}{\Gamma(1/2)}\frac d{dx} 2\sqrt x = \frac1{\sqrt{πx}}$$
The integral formula makes it clear that $D^\alpha f$ is $\mathbb R$-linear in $f$. Hence we can piece these together,
$$ D^{1/2}(D^{1/2} \sqrt{x}) = D^{1/2}\frac{\sqrt{π}}{2} = \frac{\sqrt{π}}{2}D^{1/2} 1 = \frac{\sqrt{π}}{2} \frac1{\sqrt{πx}} = \frac1{2\sqrt x} \color{green}{\overset{\checkmark}{=} \frac{d}{dx} \sqrt x}$$
There is a proof on Wikipedia that the fractional integral commutes $J^{\alpha+\beta} = J^{\alpha}J^\beta$, but I quote the following paper (which has proofs),

Although the Riemann-Liouville integral operator $D^{−α},  α ∈ \mathbb R^+ $  has the semigroup
      property[...] the RL derivative operator does not. However, we have the following interesting result. 
If $x(t)  ∈ C^1 [0,T] , α_i ∈  [0,1]  \ (i\in   1,2 ) $  and $ α_1+   α_2 ∈  (0, 1 ]$, then $D^{α_1} · D^{α_2} x( t) =    D^{α_1+ α_2} x (t) $ .

By the way, linearity implies that you know what solves $D^{1/2} f = 1$ since you computed $D^{1/2}\sqrt x $ is a constant.
