From my point of view, your definition is strange, or at least it has a contra-intuitive feature: because $x$ and $\delta$ should have the same unit (e.g. some length unit), $\delta^\alpha$ will have the unit e.g. length$^\alpha$. Then $\tilde D^a f(a)$ will have the unit "meter$^{1-\alpha}$". But maybe this is OK? Please correct me if I'm wrong.
(This was meant as a comment and not an answer. Now that I posted this as an answer, I however after deleting it cannot post a comment... Therefore I un-delete the answer, but reader please consider it as a comment :-) )
