Differentiation formulas for exterior derivative

Suppose $f,g$ are two discrete $0$-forms.

Sum Rule. $$d(f + g)([n,n+1]) = df([n,n+1]) + dg([n,n+1]).$$

Constant Multiple Rule. $$d(cf)([n,n+1]) = c\cdot df([n,n+1]).$$

So, the exterior derivative is linear, of course: $$d(\alpha f+\beta g)=\alpha df+\beta dg.$$

Product Rule. $$d(f \cdot g)([n,n+1]) = f(n + 1)dg([n,n+1]) + df([n,n+1])g(n).$$

Quotient Rule. $$d(f/g)([n,n+1]) = \frac{df([n,n+1])g(n) − f(n)dg([n,n+1])}{g(n)g(n + 1)},$$ provided $g(n),g(n+1) \ne 0$.

"Natural" Exponent Formula. $$d(2^x)([n,n+1])=2^n.$$

"Falling" Power Formula. $$d (x^{\underline {k}})([n,n+1])=kn^{\underline {k-1}},$$ where $$n^{\underline {k}} = n(n−1)(n−2)(n−3)...(n−k+1). $$

Chain Rule. For an integer-valued $0$-form $g$, $$d(fg)=fdg;$$ and $$d(fg)=dfg_1,$$ where $g_1$ a $1$-chain map generated by $g$ (Note: $g_1$ is, in a way, the derivative of $g=g_0$).

Proof. Using the Stokes Theorem with $a=[n,n+1]$ (or any other $1$-chain), we have $$d(fg)(a)=(fg)(\partial a)=f(g(\partial a)).$$ Then, in first case, using the Stokes Theorem again, we have $$=f(dg)(a).$$ In the second case, using $\partial g_*=g_*\partial$ property with $g_0=g$, we have $$=f\partial(g_1(a))=(df)(g_1(a)).\blacksquare$$