If $a\in C^\infty(\Omega)$ and $T\in \mathscr{D}'(\Omega)$ How to prove $$D^\alpha(aT)=\sum_{\beta\leq\alpha}\binom{\alpha}{\beta}D^\beta aD^{\alpha-\beta}T$$
The following is my attempt. $$ \begin{align} D^\alpha(aT)(g)&=(-1)^{|\alpha|}(aT)(D^\alpha g)\\ &=(-1)^{|\alpha|}T(aD^\alpha g) \end{align} $$ And $$ \begin{align} \sum_{\beta\leq\alpha}\binom{\alpha}{\beta}(D^\beta aD^{\alpha-\beta}T)(g)&=\sum_{\beta\leq\alpha}\binom{\alpha}{\beta}D^{\alpha-\beta}T(D^\beta a g)\\ &=\sum_{\beta\leq\alpha}\binom{\alpha}{\beta}T(D^{\alpha-\beta}(D^\beta a g)) \end{align} $$ Why are the two formulas equal ?