Read up on the Lagrange 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.

Read up on the Lagrange 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 $x^\mu$.