Double Root of Polynomial is Root of Derivative

Theorem
Let $R$ be a ring.

Let $R \left[{x}\right]$ be the ring of polynomial forms over $R$.

Let $f\in R[x]$.

Suppose that $a\in R$ is a double root of $f$.

Let $f'$ denote the formal derivative of $f$.

Then $a$ is a root of $f'$.

Proof
Because $a$ is a double root of $f$, we can write $f(x)=(x-a)^2g(x)$ with $g(x)\in R[x]$.

From Formal Derivative of Polynomials Satisfies Leibniz's Rule we have


 * $f'(x)=2(x-a)g(x)+(x-a)^2g'(x)$

and thus $f'(a)=0$.