how to prove $[D_{\beta}, D_{\alpha}]AB=([D_{\beta}, D_{\alpha}]A)B+([D_{\beta}, D_{\alpha}]B)A$?

Question

Where $A$, $B$ are arbitrary scalars, four-vectors, or tensors

we have this property: $$ [D_a, D_b]V^c=R^c_{eba}V^e $$ with $R^c _{eba}$ is the Riemann tensor

Answer 1

A straighforward calculation gives $$ (D_\beta D_\alpha) (A B) = D_\beta ((D_\alpha A) B + A (D_\alpha B)) \\ = (D_\beta D_\alpha A) B + (D_\alpha A) (D_\beta B) + (D_\beta A) (D_\alpha B) + A (D_\beta D_\alpha B) $$

By just swapping $\alpha$ and $\beta$ we have $$ (D_\alpha D_\beta) (A B) = (D_\alpha D_\beta A) B + (D_\beta A) (D_\alpha B) + (D_\alpha A) (D_\beta B) + A (D_\alpha D_\beta B) $$

Subtracting the second one from the former gives $$[D_\beta, D_\alpha](AB) = ([D_\beta, D_\alpha]A) B + A ([D_\beta, D_\alpha] B)$$

Answer 2

Presumably you're given that $D_\alpha$ and $D_\beta$ are derivations and that $[D_\beta,D_\alpha]$ means their commutator $D_\beta D_\alpha-D_\alpha D_\beta$. With all that, just write out both sides of your desired equation, apply the hypotheses that $D_\alpha$and $D_\beta$ are derivations, cancel some terms, and observe that the two sides are the same.