Product of values of the derivative of a polynomial at its roots

Question

Let $K$ be a field and $f(x) \in K[x]$ be irreducible and separable, with distinct roots $\alpha_1, \ldots, \alpha_n$. In my notes for my algebraic number theory class, I see the equality $$ \prod_{i=1}^n f'(\alpha_i) = \prod_{j \neq i} (\alpha_i - \alpha_j) $$ where $f'(x)$ denotes the formal derivative. Where does this equality come from?

Answer 1

As potla wrote, $f(x) =\prod_{i=1}^n (x-a_i) $.

To get $f'(x)$, use $\ln f(x) =\sum_{i=1}^n \ln(x-a_i) $ so $\dfrac{f'(x)}{f(x)} =\sum_{i=1}^n \dfrac1{x-a_i} $ or

$\begin{array}\\ f'(x) &=f(x)\sum_{i=1}^n \dfrac1{x-a_i}\\ &=\sum_{i=1}^n \dfrac{f(x)}{x-a_i}\\ &=\sum_{i=1}^n \prod_{j=1, j\ne i}^n(x-a_i)\\ \end{array} $

Note: If you don't like the use of $\ln$, you can prove this by induction on $n$ using the product rule.

Therefore $f'(a_k) =\sum_{i=1}^n \prod_{j=1, j\ne i}^n(a_k-a_j) = \prod_{j=1, j\ne k}^n(a_k-a_j) $ since all the other terms (with $k \ne i$) have $a_k-a_k$.

Therefore $\prod_{k=1}^n f'(a_k) = \prod_{k=1}^n \prod_{j=1, j\ne k}^n(a_k-a_j) $.