Show that $d(f.g)=f.dg+g.df$

Question

if $f,g:\mathbb{R^n \to\mathbb{R}}$ be a differentiable. then show$$d(f.g)=f.dg+g.df$$

here $f,g$ are $0$ forms on $\mathbb{R^n}$. and so is $(f.g)$ Then using theorem if $\omega_1$ is $k$ form and $\omega_2$ is $l$ form then$$d(\omega_1\land\omega_2)=d(\omega_1)\land \omega_2 + (-1)^k \omega_1 \land d(\omega_2)$$

can we apply this theorem to our question and get the answer. Else some other way ?

Thanks in advanced.

Answer 1

Absolutely. The exterior product with a $0$-form is just the product by a scalar (at each point). I.e. by definition $f \wedge \omega = \omega \wedge f = f \cdot \omega$. The formula applies, with $k=l=0$.

But note that Leibniz's rule for $0$-forms is a motivation for the construction of the exterior deriviative on the de Rham complex, rather than a corollary of it.