Lagrange polynomial second order derivative Well I came across a problem to find a generalized version ($n+1$ nodes) of first and second order derivatives for Lagrange interpolation polynomial. In some former post, I found an expression for deriving $L_j(x)$, where $L_j$ stands for Lagrange basis polynomial. The expression is as follows:
\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}
Now, what I wanted to do naturally was to find second order derivative for $L_j(x)$, and as each element in main sum has only an expression
\begin{align}
\prod_{m\not = (j,l)} {x-x_m}
\end{align}
which contains $X$, and the other expressions are constants, I need to find derivative of the above expression, but how do I do this?
 A: 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.
A: $$L_{j}(x) = \prod_{i\neq j} \frac{x-x_{i}}{x_{j}-x_{i}} $$
then $$ ln\Big(L_{j}(x)\Big) = ln\Big(\prod_{i\neq j} \frac{x-x_{i}}{x_{j}-x_{i}}  \Big) = \sum_{i \neq j } ln\Big( \frac{x-x_{i}}{x_{j}-x_{i}} \Big) $$
if we derivate we have: 
$$ \frac{L'_{j}(x)}{L_{j}(x)} =\sum_{i \neq j} \frac{\frac{1}{x_{j}-x_{i}}}{\frac{x-x_{i}}{x_{j}-x_{i}}} = \sum_{i \neq j} \frac{1}{x-x_{i}}  $$ 
then 
$$ L'_{j}(x) = L_{j}(x) \Big( \sum_{i \neq j} \frac{1}{x-x_{i}}  \Big)  $$ 
if we use the product rule for  derivatives we have that: 
$$ L''_{j}(x) = L'_{j}(x) \Big( \sum_{i \neq j} \frac{1}{x-x_{i}}  \Big)+L_{j}(x) \Big( \sum_{i \neq j} \frac{1}{x-x_{i}}  \Big)' \\ = 
 L'_{j}(x) \Big( \sum_{i \neq j} \frac{1}{x-x_{i}}  \Big)+L_{j}(x) \Big( \sum_{i \neq j} \frac{-1}{(x-x_{i})^2}  \Big) $$ we know $ L'_{j}(x)  $  so $$ \\ = 
 L_{j}(x) \Big( \sum_{i \neq j} \frac{1}{x-x_{i}}  \Big)\Big( \sum_{i \neq j} \frac{1}{x-x_{i}}  \Big)-L_{j}(x) \Big( \sum_{i \neq j} \frac{1}{(x-x_{i})^2}  \Big)\\ = 
L_{j}(x) \Big( \sum_{i \neq j} \frac{1}{x-x_{i}}  \Big)^2 -L_{j}(x) \Big( \sum_{i \neq j} \frac{1}{(x-x_{i})^2}  \Big) \\ = 
L_{j}(x) \Big\{\Big( \sum_{i \neq j} \frac{1}{x-x_{i}}  \Big)^2 -  \sum_{i \neq j} \frac{1}{(x-x_{i})^2}     \Big\} $$
