Let $d_c, \delta_c$ be operators with domains $D(d_c) = D(\delta_c) = C_{c}^\infty(\wedge T^\ast M)$. We let $d_c$ be the usual exterior derivative on compactly supported smooth forms, ie., $d_c\omega = dx^k \wedge \nabla_k \omega$ and $\delta_c = dx^k \llcorner \nabla_k \omega$.

We define $d:D(d) \subset L^2 \to L^2$ as the adjoint of $\delta_c$. Ie, the operator with largest domain $D(d)$ satisfying $(d\omega,\eta) = (\omega,\delta_c \eta)$ for all $\omega \in D(d)$ and $\eta \in C_{c}^\infty$.

How do I show that for some smooth coordinate chart $\phi:U \subset M \to \phi(U) \subset \mathbb{R}^n$ that $d(\phi^{-1})^\ast\omega = (\phi^{-1})^\ast d\omega$ for $\omega \in D(d)?$

Note that I don't know that $\bar{d_c}= d$, so I can't simply approximate by smooth forms. In fact, this arises in trying to prove exactly that statement.