We define for example $H^{1/2}(\mathbb R)$ as the set $$\left\{f\in L^2(\mathbb R)\mid \int_{\mathbb R}(1+|\xi|^2)^{1/2}|\hat f(\xi)|\mathrm d \xi<\infty \right\},$$ where $\hat f$ is the Fourier transform of $f$. As well, you can see here that we can define $$\frac{d^{1/2}}{dx^{1/2}}f(x)=\frac{1}{\Gamma(1/2)}\int_0^x(x-t)^{-1/2}f(t)dt$$ when $f$ has sufficiently good condition (we can also generalize this for $\frac{d^\alpha }{dx^\alpha } f$ for $\alpha >0$).
My question is : we don't have explicitly the $1/2-$derivative of $f$ when it's in $H^{1/2}$ but if we can catch it, and if the function is regular enough, would it be the same derivative ?