Along Demailly's agbook, let $\mathscr R$ be a sheaf of commutative rings and $\mathscr A, \mathscr B$ sheaves of $\mathscr R$-modules. For a sheaf $\mathscr F, \mathscr (F^{\bullet}, d)$ denotes the simplicial flasque resolution of $\mathscr F$.
Here $d^q$ has been defined by $$d^qf = \sum_{0\leq j\leq q}(-1)^j f(x_0,\dots,\hat{x_j},\dots,x_{q+1}) + f(x_0,\dots,x_q)(x_{q+1}) \tag{D}$$
If $f\in\mathscr A^{[p]}_{x}, g\in\mathscr B^{[q]}_{x}$, define the cup product as $$f\smile g = f(x_0,\dots, x_p)(x_{p+q})\otimes g(x_p,\dots,x_{p+q}) \in (\mathscr A \otimes_\mathscr R\mathscr B)_x^{[p+q]}.$$
I want to show that $$d^{p+q}(f\smile g) = (d^p f) \smile g + (-1)^pf\smile (d^q g).$$
So I first calculated both parts of the right side: $$(1) = (d^pf)\smile g \\= \left(\sum_{0\leq j \leq p}(-1)^j f(x_0,\dots,\hat{x_j},\dots,x_{p+1}) + (-1)^{p+1}f(x_0,\dots,x_p)(x_{p+1})\right)(x_{p+q+1})\otimes g(x_{p+1},\dots,x_{p+q+1}),$$
$$ (2) = f\smile (d^q)g \\ = f(x_0,\dots,x_p)(x_{p+q+1})\otimes (-1)^{-p}\left(\sum_{p\leq j\leq p+q}(-1)^jg(x_p,\dots,\hat{x_j},\dots,x_{p+q+1})+(-1)^{p+q+1}g(x_p,\dots,x_{p+q})(x_{p+q+1})\right) $$
Now $(1)+(-1)^p(2)$ comes sort of close to the definition of $d^{p+q}(f\smile g)$ by $(D)$, but I fail to see the equality. Can someone point out my mistakes?