I may be missing something simple here, but in the section for stability of Wikinson's polynomial in Wikipedia the derivative is defined as:
$-\frac{\alpha^{19}}{\prod_{k \ne j}(\alpha_j - \alpha_k)}$
The denominator is defined as $p^{\prime}(\alpha_j)$ but I don't see where this derivative comes from? I can see it has to be something to do with the product rule which sets to zero all the non-k derivatives, but I am not entirely sure which variables are being changed here.
The general idea is that you have a polynomial $p$ with one exact root $\alpha$. Due to numerical representation of the coefficients or the evaluation procedure the computer sees another polynomial that can, to some degree, be represented as $p(x)+\epsilon c(x)$. The basic idea of the perturbation calculation is now that this new polynomial still has a root close to $\alpha$, write it as $\alpha+\delta$. Now look at the Taylor expansion $$ 0+p'(\alpha)\delta+O(\delta^2)+\epsilon c(\alpha)+O(\epsilon\delta) $$ If $p'(\alpha)$ is large against $\epsilon$ and $\delta$, then a root approximation for the perturbed polynomial can be found at $$\delta=-\epsilon\frac{c(\alpha)}{p'(\alpha)},$$ that is $$ x=\alpha-\epsilon\frac{c(\alpha)}{p'(\alpha)} $$ If you see that as the start of a Taylor expansion of the root curve $x(\epsilon)$, then this second term is the linear term of the expansion associate with the derivative $x'(0)$.
The derivative of $p(x)=\prod_{k=1}^n(x-\alpha_k)$ is $$ p'(\alpha_m)=\lim_{x\to\alpha_m}\frac{p(x)-0}{x-\alpha_m}=\prod_{k\ne m}(\alpha_m-\alpha_k). $$
