Let me just quickly provide the reference to the original question (It was posted by me and I had some problems to derive the formula: Link)
So we start with
\begin{align}
L_j'(x) = \sum_{l\not = j} \frac{1}{x_j-x_l}\prod_{m\not = (j,l)} \frac{x-x_m}{x_j-x_m}
\end{align}
as you already mentioned, the interesting part is the product
\begin{align}
\prod_{m\not =(j,l)} \frac{x-x_m}{x_j-x_m}
\end{align}
To compute the derivative, we apply the chain rule to get
\begin{align}
\frac{d}{dx}\left(
\prod_{m\not =(j,l)} \frac{x-x_m}{x_j-x_m}
\right)
=
\sum_{w \not = (j,l)} \frac{1}{x_j-x_w}\prod_{m\not = (j,l,w)}\frac{x-x_m}{x_j-x_m}
\end{align}
Putting this together, we end up with
\begin{align}
L_j''(x) = \sum_{l\not = j} \frac{1}{x_j-x_l}
\sum_{w \not = (j,l)}\frac{1}{x_j-x_w} \prod_{m\not = (j,l,w)}\frac{x-x_m}{x_j-x_m}
\end{align}
Like in the last thread, I'm sure that this can be simplified, but since I failed the last time, I don't want to attempt this again.
Applying method of differences to $p(0), p(1)$, we see that the polynomial must be of the form $ p(x) = (x+ 1) + x(x-1) \times A(x)$.
Checking the condition at $p'(2)$, we get $ - \frac{1}{2} = p'(2) = 1 + ( 4 - 1 ) \times A(x) + 2(2-1) \times A'(x) $.
So, we could have $A(x) = - \frac{1}{2}$, which yields $ - \frac{x^2}{2} + \frac{3x}{2} + 1 $.
Best Answer
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 $$ p'(\alpha_m)=\lim_{x\to\alpha_m}\frac{p(x)-0}{x-\alpha_m}=\prod_{k\ne m}(\alpha_m-\alpha_k). $$