Restating my comment (after looking at the formulas a second time), I am highly surprised if it's not the **Leibniz rule**, which Oscar rewords as a graded derivation. Sketch of proof: Cup product works on mixed relative cohomologies, $H^p(X,A_1)\times H^q(X,A_2)\to H^{p+q}(X,A_1\cup A_2)$, so we can take $A_1=A_2=A$ or $A_1=A_2=\varnothing$ or $A_1=A$ and $A_2=\varnothing$. It's really just definition-checking from here on out. The coboundary map $H^n(A)\to H^{n+1}(X)$ is obtained by taking co-chains on $A$ and viewing them as co-chains on $X$ which vanish on $X-A$ and then pre-composing with the differential $C_{n+1}(X)\to C_n(X)$. If you work out the formula for the co-differential of the cup product of co-chains (which is the Leibniz rule), this should respect the values of the relative co-chains. And the coboundary map in the long exact sequence for relative cohomology is obtained directly from the co-differential. So in your notation, $\partial_{p+q}(\alpha\cup\beta) = \partial_p\alpha\cup\beta\pm \alpha\cup\partial_q\beta$, and I'll leave you to the signs (I think that's a phrase I hear way too often in math).