Let $x$, $y$ be 0-forms, thus $dx$, $dy$ are 1-forms. Since 1-forms compose an algebra over 0-forms ring, expressions like
$$y dx$$
make perfect sense. Now I ask what is
$$d(y dx)$$
I suggest it to be $y d(dx) = 0$, since $d$ is linear, however I feel it is likely to be wrong. Is there any other meaningful product except $\land$ between forms that would generalize the situation above?
$dƒ$ is the differential of $ƒ$ for smooth functions $ƒ$.
$d(dƒ) = 0$ for any smooth function $ƒ$.
$d(α∧β) = dα∧β + (−1)^p(α∧dβ) $ where $\alpha$ is a p-form. Take $\alpha=y$ and $\beta=dx$ in this last formula.