Half derivate of x

$\frac{\partial^{1/2} }{\partial x^{1/2}}x$

Let's start with:

$f(a)=x^{k}$

$f(a)'=kx^{k-1}$

In general:
$\frac{\partial^{a} }{\partial x^{a}}x^{k}$ = $\frac{k!}{(k-a)!}$ $x^{k-a}$

Using Gamma Function:

$\frac{\partial^{a} }{\partial x^{a}}x^{k}$ = $\frac{\Gamma (k+1)}{\Gamma (k-a+1)}$ $x^{k-a}$
$k\geq 0$

For k=1 and a=1/2, we obtain:

$\frac{\partial^{1/2} }{\partial x^{1/2}}x$ = $\frac{\Gamma (1+1)}{\Gamma (1-\frac{1}{2}+1)} x^{1-\frac{1}{2}}= \frac{\Gamma (2)}{\Gamma (\frac{3}{2})} x^{\frac{1}{2}}= \frac{1}{\frac{\pi^{\frac{1}{2}} }{2}} x^{\frac{1}{2}}$

$\frac{\partial^{1/2} }{\partial x^{1/2}}x$ = $\frac{2x^{\frac{1}{2}}}{\pi^{\frac{1}{2}} }$

Now, derivating for "Half derivative" to demonstrate:

$\frac{\partial^{1/2} }{\partial x^{1/2}}x$ $\frac{2x^{\frac{1}{2}}}{\pi^{\frac{1}{2}} }$ = $\frac{2}{\pi ^{\frac{1}{2}}} \frac{\Gamma (1+\frac{1}{2})}{\Gamma (\frac{1}{2}-\frac{1}{2}+1)} x^{\frac{1}{2}-\frac{1}{2}}= \frac{2}{\pi ^{\frac{1}{2}}} \frac{\Gamma (\frac{3}{2})}{\Gamma (1)} x^{0}= \frac{2\frac{\pi ^{\frac{1}{2}}}{2}}{\pi \frac{1}{2}} x^{0}=1$

So:

$(\frac{\partial^{1/2} }{\partial x^{1/2}} \frac{\partial^{1/2} }{\partial x^{1/2}})x= \frac{\partial }{\partial x}x=1$