Just to complement the answer, here is the formula for second derivative:
$$
L^{"}_i(x) =\sum_{l\ne i}\frac{1}{x_i-x_l}\left( \sum_{m\ne(i,l)}\frac{1}{x_i-x_m}\prod_{k\ne(i,l,m)}\frac{x-x_k}{x_i-x_k}\right)
$$
through recursion, one can compute further higher derivatives.
In short, this is product rule. If you are comfortable with the notation for the product rule of $N$ factors (see e.g. Wikipedia), denoted by "$(\text{PR})$" below, you may calculate:
$$\begin{align*}D_x[(x-x_0)\ldots(x-x_n)]\rvert_{x=x_j}&=D_x[\prod_{i=1}^n (x-x_i)]\rvert_{x=x_j}\\
&\overset{\text{(PR)}}{=}\sum_{i=0}^n\Bigg( \underbrace{D_x[x-x_i]\vphantom{\prod_{k=0,k\neq i}^n}}_{=1}\prod_{k=0,k\neq i}^n (x-x_k)\Bigg)\Bigg\rvert_{x=x_j}\\
&=\sum_{i=0}^n\underbrace{\prod_{k=0,k\neq i}^n \underbrace{(x_j-x_k)}_{=0\text{ for }k=j}}_{=0 \text{ for }i\neq j}\\
&=\prod_{k=0,k\neq j}^n (x_j-x_k)
\end{align*}$$
Step-by-step approach may look like this:
I guess what happens to the left and the right term in your 2nd display line is clear, right? So we know that
$$\begin{align*}&f'(x_j)=\\
&=\sum_{k=0}^n f(x_k)L_k'(x_j)+\left.\left(D_x[\tfrac{(x-x_0)\ldots(x-x_n)}{(n+1)!}]\right)\right\rvert_{x=x_j}f^{(n+1)}(\epsilon(x))+\underbrace{\tfrac{(x-x_0)\ldots(x-x_n)}{(n+1)!}}_{=0}\left(\left.D_x[f^{(n+1)}(\epsilon(x))]\right)\right\rvert_{x=x_j}\\
&=\sum_{k=0}^n f(x_k)L_k'(x_j)+\left.\left(D_x[\tfrac{(x-x_0)\ldots(x-x_n)}{(n+1)!}]\right)\right\rvert_{x=x_j}f^{(n+1)}(\epsilon(x))\end{align*}$$
and your question is about the remaining term without the $L_k$'s. We notice that $\left.\left(D_x[\tfrac{(x-x_0)\ldots(x-x_n)}{(n+1)!}]\right)\right\rvert_{x=x_j}=\tfrac{1}{(n+1)!}\left.\left(D_x[(x-x_0)\ldots(x-x_n)]\right)\right\rvert_{x=x_j}$ and calculate the derivative by using product rule:
$$\begin{align*}&\left.\left(D_x[(x-x_0)\ldots(x-x_n)]\right)\right\rvert_{x=x_j}\\
\\
=&\left.\bigg(\underbrace{D_x[x-x_0]}_{=1}(x-x_1)(x-x_2)\ldots(x-x_n)\bigg)\right\vert_{x=x_j}+\\
&\left.\bigg((x-x_0)\underbrace{D_x[x-x_1]}_{=1}(x-x_2)\ldots(x-x_n)\bigg)\right\vert_{x=x_j}+\\
&\qquad\ldots\qquad+\\
&\left.\bigg((x-x_0)\ldots(x-x_{n-2})(x-x_{n-1})\underbrace{D_x[x-x_n]}_{=1}\bigg)\right\vert_{x=x_j}\\
\\
=&\bigg(1\cdot(x_j-x_1)(x_j-x_2)\ldots(x_j-x_n)\bigg)+\\
&\bigg((x_j-x_0)\cdot 1\cdot(x_j-x_2)\ldots(x_j-x_n)\bigg)+\\
&\qquad\ldots\qquad+\\
&\bigg((x_j-x_0)\ldots(x_j-x_{n-2})(x_j-x_{n-1})\cdot 1\bigg)\end{align*}$$
If we fix a $j\in \{0,\ldots,n\}$, we observe that all these products contain the factor $(x_j-x_j)$ somewhere (and thus vanish) - except one! The $j$-th product (i.e. the $j$-th summand in the sum above) survives because there's a $1$ where all the other ones have the $(x_j-x_j)$. Thus
$$\begin{align*}&\left.\left(D_x[(x-x_0)\ldots(x-x_n)]\right)\right\rvert_{x=x_j}\\
=&\bigg((x_j-x_0)\ldots(x_j-x_{j-1})\cdot 1\cdot(x_j-x_{j+1})\ldots(x_j-x_n)\bigg)\\
=&\prod_{k=0,k\neq j}^n (x_j-x_k)\end{align*}$$
and consequently
$$\left.\left(D_x[\tfrac{(x-x_0)\ldots(x-x_n)}{(n+1)!}]\right)\right\rvert_{x=x_j}=\frac{1}{(n+1)!}\prod_{k=0,k\neq j}^n (x_j-x_k)$$
which proves
$$f'(x_j) = \sum_{k=0}^{n}f(x_k)L'_k(x_j) + \frac{f^{(n+1)}(\epsilon(x_j))}{(n+1)!} \prod_{k=0, k \neq j}^{n}(x_j-x_k).$$
Best Answer
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.