I'm sure you're aware that $\frac{d^n}{dx^n}\frac{1}{x}=\frac{(-1)^nn!}{x^{n+1}}$
Well, what if $n=\frac{1}{2}$?
$$\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}\frac{1}{x}=\frac{(-1)^\frac{1}{2}(\frac{1}{2})!}{x^(\frac{1}{2}+1)}$$ $$\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}\frac{1}{x}=\frac{\sqrt{-1}(\frac{1}{2})!}{x^\frac{3}{2}}$$ So would it be correct to say that: $$\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}\frac{1}{x}=\frac{i\sqrt{\pi}}{x^\frac{3}{2}}$$ Or am I just an idiot? Please tell me. :)
You may be interested in https://en.wikipedia.org/wiki/Fractional_calculus; see also Half order derivative of $ {1 \over 1-x }$ and the answer to half-derivative of $x^2$.