De Rham derivative satisfies Leibniz rule

Statement

Let $M$ be a differential manifold. Consider the de Rham derivative on $M$, defined as a map:

$d:C^{\infty }(M)\to \Gamma (T^{*}M)$

Then $d$ satisfies the Leibniz rule:

$d(fg)=g(df)+f(dg)$

In other words, $d$ is a derivation from the algebra $C^{\infty }(M)$ to the module $\Gamma (T^{*}M)$.

Proof

To verify equality of the two sides, we need to verify that for any vector field $X$, the two sides evaluate to the same thing for $X$. The left side, evaluated on $X$, yields:

$X(fg)$

The right side, evaluated on $X$, yields:

$g(Xf)+f(Xg)$

The equality of these two sides follows from the Leibniz rule for derivations.