From definition of divided differences we have that $$f[x_0,\cdots,x_n]=\sum_{j=0}^n\frac{f(x_j)}{\Pi_{{k\in\{0,\cdots,n\}-\{j\}}}(x_j-x_k)} $$ It makes completely sense to have $k\neq j$ otherwise the denominator becomes zero. In many articles I've seen the following notation, let $q(\xi)=(\xi -x_0)\cdots(\xi-x_n)$, then we can re-express the expanded form of the divided difference as $$f[x_0,\cdots,x_n]=\sum_{j=0}^n\frac{f(x_j)}{q'(x_j)}$$
In this link there's a brief explanation but I don't really get the point. why is it that $$q'(x_j)=(x_j - x_0)\cdots(x_j -x_{j-1})(x_j-x_{j+1})\cdots(x_j - x_n)$$ ? Maybe I am messing up things, but if I apply the product rule for derivatives I don't obtain that result.
Thanks in advance
First, look at the following equivalent expression of $q$: $$q(\xi)=(\xi-x_j)\cdot \prod_{i\neq j}{(\xi - x_i)}$$ Take the derivative of $q(\xi)$ with respect to $\xi$, using the product rule (no need to expand the rightmost expression): $$q'(\xi)=1\cdot \prod_{i\neq j}{(\xi - x_i)} + (\xi-x_j)\cdot \left( \prod_{i\neq j}{(\xi - x_i)}\right )'$$ Now substitute $\xi=x_j$, to get $$q'(x_j)=\prod_{i\neq j}{(x_j - x_i)} + (x_j-x_j)\cdot \left( \prod_{i\neq j}{(x_j - x_i)}\right )' = \prod_{i\neq j}{(x_j - x_i)}$$
I don't think you should view it as anything more meaningful than a shorthand for the longer expression on the right hand side.