# The Formal Derivative of a Polynomial

Recall from the Simple and Multiple Roots of Polynomials page that if $K$ is a field, $f(x) \in K[x]$, and $F$ is a splitting field of $f(x)$ over $K$ then we can write $f(x)$ as a splitting of linear factors:

(1)Where $r_1, r_2, ..., r_n \in F$ are the roots of $f(x)$ and $m_1, m_2, ..., m_n \in \mathbb{N}$ are the multiplicities of the roots. We said that a root of $f(x)$ is a simple root if it has multiplicity $1$, and a multiple root if it has multiplicity greater than or equal to $2$.

We now define the formal derivative of a polynomial.

Definition: Let $K$ be a field and let $f(x) \in K[x]$ with $f(x) = a_0 + a_1x + ... + a_nx^n$. The Formal Derivative of $f(x)$ is the polynomial $f'(x) = a_1 + 2a_2x + ... + na_nx^{n-1}$ where for each $k \in \{ 1, 2, ..., n \}$ we define $ka_k := (\underbrace{a_k + a_k + ... + a_k}_{k\:\mathrm{times}})$. |

The following theorem tells us that the sum, difference, and product rule hold for the formal derivatives of polynomials.

Theorem 1: Let $K$ be a field and let $f(x), g(x) \in K[x]$. Then:a) $[f(x) + g(x)]' = f'(x) + g'(x)$.b) $[f(x) - g(x)]' = f'(x) - g'(x)$.c) $[f(x)g(x)]' = f(x)g'(x) + f'(x)g(x)$. |

*We will only prove (a). The remaining proofs are very similar.*

**Proof of a)**Let $f(x) = a_0 + a_1x + ... + a_nx^n$ and let $g(x) = b_0 + b_1x + ... + b_mx^m$. Without loss of generality, assume that $n \geq m$. Then:

- Taking the formal derivative yields:

The following theorem generalizes an important result regarding multiple roots of polynomials.

Theorem 2: Let $K$ be a field and let $f(x) \in K[x]$. Then $f(x)$ has no multiple roots if and only if $\mathrm{gcd}(f(x), f'(x)) = 1$. |