Differential Geometry-Wedge product

How can we prove the following relation for differentiating the wedge product of a p-form $\alpha_p$ and a q-form $\beta_q$$$d(\alpha_p\wedge\beta _q)=d\alpha_p\wedge\beta_q+(-1)^{p}\alpha_p\wedge d\beta_q$$

Solution 1:

It suffices to take $\alpha = f\, dx^{i_1}\wedge\dots\wedge dx^{i_p}$ and $\beta=g\,dx^{j_1}\wedge \dots\wedge dx^{j_q}$. Then $\alpha\wedge\beta = fg\, dx^{i_1}\wedge\dots\wedge dx^{i_p}\wedge dx^{j_1}\wedge \dots\wedge dx^{j_q}$. Write out the product rule, and then keep track of the switches you must do to get $d\alpha\wedge\beta$ and $\alpha\wedge d\beta$ appearing in your formula.

EDIT: Doing the exercise for you: \begin{align*} d(\alpha\wedge\beta) &= d(fg\, dx^{i_1}\wedge\dots\wedge dx^{i_p}\wedge dx^{j_1}\wedge \dots\wedge dx^{j_q}) \\ &= (g\,df + f\,dg) \wedge (dx^{i_1}\wedge\dots\wedge dx^{i_p}\wedge dx^{j_1}\wedge \dots\wedge dx^{j_q}) \\ &= (df\wedge dx^{i_1}\wedge\dots\wedge dx^{i_p})\wedge (g\,dx^{j_1}\wedge \dots\wedge dx^{j_q}) + \\ &\qquad (f\,dg)\wedge (dx^{i_1}\wedge\dots\wedge dx^{i_p}\wedge dx^{j_1}\wedge \dots\wedge dx^{j_q}) \\ &= d\alpha \wedge\beta + (-1)^p (f\,dx^{i_1}\wedge\dots\wedge dx^{i_p})\wedge (dg\wedge dx^{j_1}\wedge \dots\wedge dx^{j_q}) \\ &= d\alpha\wedge\beta + (-1)^p \alpha\wedge d\beta. \end{align*}