Question on theorem 10.20 Walter Rudin [duplicate]

Andy, briefly, any $q$-form can be written as $$\sum_{j_1<j_2<\dots<j_q} f_{j_1j_2\dots j_q}dx_{j_1}\wedge\dots\wedge dx_{j_q} = \sum_J f_J dx_J,$$ for short. Since $\wedge$ is bilinear and since the exterior derivative of a sum is the sum of the exterior derivatives, it suffices to take just one such term for each of $a$ and $b$ and take $$a = f_J\,dx_J \quad\text{and}\quad b = g_I\,dx_I.$$ Then $a\wedge b = f_Jg_I dx_J\wedge dx_I$ (which will be $0$, of course if there's repetition in the multiindices). Now use the product rule and rearrange terms (see note below) to get your answer. (Note that if there's repetition in these multiindices, that repetition will persist when you differentiate and so you'll get $0$ automatically on the right-hand side, too.)

(Note: $$dx_r\wedge dx_J\wedge dx_I=(-1)^q dx_J\wedge dx_r\wedge dx_I$$ for any $1\leq r\leq n$.)