Half order derivative of $ {1 \over 1-x }$ I'm new to this "fractional derivative" concept and try, using wikipedia, to solve a problem with the half-derivative of the zeta at zero, in this instance with the help of the zeta's Laurent-expansion.      
Part of this fiddling is now to find the half-derivative  $$ {d^{1/2}\over dx^{1/2}}{1 \over 1-x}$$
First I would like to understand, whether there is a short/closed form for this ot whether I have to express the fraction as a power series first and then to differentiate termwise.   
Next I would like to know the value at $x=0$.     
 A: $$\frac{d^k}{dx^k} \left(\frac{1}{1-x} \right)=\frac{d^k}{dx^k} (1+x+x^2+...)=\Gamma(k+1) \sum_{n=k}^{\infty} {n\choose k}x^{n-k}\\
\sum_{n=k}^{\infty} {n\choose k}x^{n-k}=\frac{1}{(1-x)^{k+1}}\\
\frac{d^k}{dx^k} \frac{1}{1-x}=\frac{\Gamma(k+1)}{(1-x)^{k+1}}$$
A: See Reference  for a complete solution of the problem of differentiation and integration of real order of rational polynomials.  
Formal approaches to fractional derivative: There are several definitions for Fractional derivative. The most widely known one is the Riemann-Liouville fractional derivative
$$ f^{(q)}(x) = \frac{1}{\Gamma(k-q)} \frac{d^k}{dx^k} \int_{a}^{x}\,
(x-t)^{k-q-1}\,f(t)\,dt\>, \quad (k-1 < q < k )\,, $$
and according to this definition you will have the following answer
$$ f^{\left(\frac{1}{2}\right)}(x)={\frac {{_2F_1\left(1,1;\,\frac{1}{2};\,x\right)}}{\sqrt {x}\sqrt {\pi }}}, $$
where $ _2F_1 $ is the hypergeometric function. The limit of the above function as $x\to 0^+$ goes to infinity. However, there is another definition known as the Caputo definition and is defined as
$$ f^{(q)}(x) = \frac{1}{\Gamma(k-q)}  \int_{a}^{x}\,
(x-t)^{k-q-1}\,\frac{d^k}{dt^k}f(t)\,dt\>, \quad (k-1 < q < k )\,, $$
and this definition will give you a different answer, namely
$$ f^{\left(\frac{1}{2}\right)}(x) = {\frac {1}{\sqrt {\pi }  \left( x-1
 \right) ^{3/2}}} \left( {\it \rm arctanh} \left( {\frac {\sqrt {x}}{
\sqrt {x-1}}} \right) -\sqrt {x}\sqrt {x-1}\right), $$
and the limit in this case goes to $0$.
Note: The simplest approach to your problem is summarized in the steps 
1) Compute the Taylor series of the function,
2) Use the Formula
$$ \frac{d^q}{dx^q} x^m = \frac{\Gamma(m+1)}{\Gamma(m-q+1 )} x^{m-q}\,, $$
which corresponds to the Riemann-Liouville definition of the function $x^m$, or the formula (2.29), page 15 which corresponds to Caputo definition. 
See here to see the power series technique.
A: Use what you know about whole number derivatives. Inductively, you can prove $$\frac{d^n}{dx^n}\frac1{1-x}=\frac{n!}{(1-x)^{n+1}}$$ Now express $n!$ using the $\Gamma$ function ($\Gamma(n+1)$), and you can extend the definition to non-integral $n$: $$\frac{d^{1/2}}{dx^{1/2}}\frac1{1-x}=\frac{\Gamma(3/2)}{(1-x)^{3/2}}$$ At $x=0$, this is just $\Gamma(3/2)$.

To confirm that this method works, observe that you can also inductively prove $$\begin{align}\frac{d^n}{dx^n}\frac1{(1-x)^{3/2}}&=\frac{\frac{(2n+1)!}{4^n\cdot n!}}{(1-x)^{n+3/2}}\\&=\frac{\frac{\Gamma(2n+2)}{4^n\Gamma(n+1)}}{(1-x)^{n+3/2}}\end{align}$$ and extend to nonintegral $n$, so that $$\begin{align}\frac{d^{1/2}}{dx^{1/2}}\frac{d^{1/2}}{dx^{1/2}}\frac1{1-x}&=\frac{d^{1/2}}{dx^{1/2}}\frac{\Gamma(3/2)}{(1-x)^{3/2}}\\&=\frac{\Gamma(3/2)\frac{\Gamma(3)}{2\Gamma(3/2)}}{(1-x)^{2}}\\&=\frac{\frac{2!}{2}}{(1-x)^{2}}\\&=\frac{1}{(1-x)^2}\\&=\frac{d}{dx}\frac{1}{1-x}\end{align}$$ and all is as it should be.
A: While @alex.jordan's answer was nice and all, it fails for $-n\in\mathbb N$, where the logarithm should come into play.  To account for such behavior, one could use the following formula:
$$\frac{d^n}{dx^n}\ln(x)=\frac{\ln(x)-\gamma-\psi^{(0)}(1-n)}{x^n\Gamma(1-n)}$$
which may also be proven by induction or derived straight from formulas.  Here we have the Euler-Mascheroni constant and the digamma function.
Changing the argument to $1-x$, we have
$$\frac{d^n}{dx^n}\ln(1-x)=\frac{\ln(1-x)-\gamma-\psi^{(0)}(1-n)}{(x-1)^n\Gamma(1-n)}$$
where we applied chain rule and put it in the denominator.
$$\frac{d^{n+1}}{dx^{n+1}}\ln(1-x)=\frac{\ln(1-x)-\gamma-\psi^{(0)}(-n)}{(x-1)^{n+1}\Gamma(-n)}$$
$$\frac{d^n}{dx^n}\frac{-1}{1-x}=\frac{\ln(1-x)-\gamma-\psi^{(0)}(-n)}{(x-1)^{n+1}\Gamma(-n)}$$
$$\frac{d^n}{dx^n}\frac1{1-x}=\frac{\ln(1-x)-\gamma-\psi^{(0)}(-n)}{-(x-1)^{n+1}\Gamma(-n)}$$
I will remark the only problem with this is with $n\in\mathbb N$, where will get the indeterminate form $\frac\infty\infty$, which can be treated as a limit to get the desired values.  Also note that by applying L'Hospital's rule with respect to $n$ removes the logarithm for $n\in\mathbb N$, which explains why we usually do not see it occur.
Lastly, the half derivative is given as
$$\frac{d^{1/2}}{dx^{1/2}}\frac1{1-x}=\frac{\ln(1-x)-\gamma-\psi^{(0)}(-1/2)}{-(x-1)^{3/2}\Gamma(-1/2)}$$
$$=\frac{\ln(1-x)+\ln(4)}{2\sqrt\pi(x-1)^{3/2}}$$
And at $x=0$,
$$=\frac{-i\ln(2)}{\sqrt\pi}$$
where $i=\sqrt{-1}$
