Formal Derivative of Polynomials Satisfies Leibniz's Rule

Theorem
Let $R$ be a commutative ring with unity.

Let $R \sqbrk X$ be the polynomial ring over $R$.

Let $f, g \in R \sqbrk X$ be polynomials.

Let $f'$ and $g'$ denote their formal derivatives.

Then:
 * $\paren {f g}' = f g' + f' g$

Proof
Both sides are bilinear functions of $f$ and $g$, so it suffices to verify the equality in the case where $\map f X = X^n$ and $\map g x = X^m$.

Then:
 * $\paren {X^n X^m}' = \paren {n + m} X^{n + m - 1}$

and:
 * $\paren {X^n}' X^m + X^n \paren {X^m}' = n X^{n - 1} X^m + m X^n X^{m - 1}$