Read up on the [Lagrange shift operator](https://en.wikipedia.org/wiki/Shift_operator), essentially a formal summary of the Taylor expansion. You know its result given your summary of translations. 

For just one variable, you found
$$
e^{a\partial_x} f(x) = f(x+a).
$$

Now note for $y\equiv \ln x$,
$$
e^{ax \partial_x} f(x) ~~~\leadsto \\
e^{a \partial_y} f(e^y) =f(e^{y+a})=f(e^a~ x),
$$
so your $\alpha=e^a$. 

You can take it from here, appreciating *D* is scaling only the rotationally invariant "radius" of $x^\mu$.