Perturbation of roots in Wilkinson's polynomial I am studying numerical analysis. When I read the online definition I found on this paragraph: * Suppose that we perturb a polynomial $p(x) = Π (x−α_j)$ with roots $α_j$ by adding a small multiple $t·c(x)$ of a polynomial $c(x)$, and ask how this affects the roots $α_j$. To first order, the change in the roots will be controlled by the derivative $\frac{d\alpha_j}{dt}=-\frac{c(\alpha_j)}{p'(\alpha_j)},p'(\alpha_j)$ is the derivative of $p(x)$. How did they get the formula for the change of the root=$-\frac{c(\alpha_j)}{p'(\alpha_j)}$? This same definition also appears in Trefethen's numerical linear algebra book, but there's no explanation there either. I know this may be a dumb question but thanks for helping me out!!

Define $F(x, t) = p(x) + tc(x)$. Then $$ F(\alpha_j, 0) = 0 \, , \quad F_x(\alpha_j, 0) = p'(\alpha_j) \
e 0 \, . $$ It follows from the implicit function theorem that there is a differentiable function $\alpha$, defined in a neighbourhood of $t = 0$, with $\alpha(0) = \alpha_j$, such that the solutions of $F(x, t) = 0$ in a neighbourhood of $(\alpha_j, 0)$ are given by $x = \alpha(t)$.

$\alpha(t)$ is the zero of the perturbed polynomial $p(x) + tc(x)$, and the derivative can be determined by differentiating $$ p(\alpha(t)) + tc(\alpha(t)) = 0 $$ with respect to $t$: $$ p'(\alpha(t)) \alpha'(t) + c(\alpha(t)) + t c'(\alpha(t))\alpha'(t)= 0 $$ Now set $t = 0$: $$ p'(\alpha_j) \alpha'(0) + c(\alpha_j) = 0 $$