Show that $\dfrac{\partial}{\partial x_i} f[x_0,x_1,\dots,x_n]=f[x_0,x_1,\dots,x_i,x_i,x_{i+1},\dots,x_n].$ Let $x_0<x_1<\cdots<x_n$, and let $f$ be continuously differentiable. Show that $$\dfrac{\partial}{\partial x_i} f[x_0,x_1,\dots,x_n]=f[x_0,x_1,\dots,x_i,x_i,x_{i+1},\dots,x_n].$$
At first, I thought this question can be done by induction on $n$ but after attempting it, I got something weird.
Next, I tried using $$f[x_0,,x_1,\dots,x_n]=\sum_{i=0}^n f(x_i) \prod_{\substack{j=0 \\ j\neq i}}^n (x_i-x_j)^{-1}$$
 But when you take the derivative of this equation. I will obtain a double sum because of the product rule. 
Can someone help me get started on this question?
 A: It is a matter of definitions. When arguments are distinct you have by the 
very definition of the divided differences, and using permutation symmetry of the objects: $$ f[x_0,...,x_{i-1},u,v,x_{i+1},..,x_n] =
    \frac{ f[x_0,...x_{i-1},u,x_{i+1}...,x_n] - f[x_0,...,x_{i-1},v,x_{i+1}...,x_n]}
{u-v}$$ 
The case when arguments are equal are defined by taking limits (not described in the wiki page as far as I can see). Your function has to be differentiable in that case. As an example:
$$ f'(x_0) =  \lim_{x_1\rightarrow x_0} \frac{f(x_1)-f(x_0)}{x_1-x_0} =
 \lim_{x_1\rightarrow x_0} f[x_0,x_1]=:f[x_0,x_0]$$
In your case:
$$ f[x_0,...,x_i,x_i,...,x_n] = \lim_{\epsilon\rightarrow 0^+}
  { f[x_0,...,x_i,x_i+\epsilon,...,x_n]} $$
Using the first part we may rewrite as: 
$$  
\lim_{\epsilon\rightarrow 0^+}
  \frac{ f[x_0,...x_{i-1},x_i+\epsilon,x_{i+1}...,x_n] - f[x_0,...,x_{i-1},x_i,x_{i+1}...,x_n]}
{(x_i+\epsilon) -x_i}  $$
which equals the wanted partial derivative (assuming it exists).
A: There is something wrong. Some derivatives of $f$ should appear. 
$$f[x_0,x_1]=\frac{f(x_0)}{x_0-x_1}+\frac{f(x_1)}{x_1-x_0}$$ and 
so
$$\frac{\partial f[x_0,x_1]}{\partial x_1}=\frac{f(x_0)}{(x_0-x_1)^2}+\frac{f'(x_1)}{x_1-x_0}-\frac{f(x_1)}{(x_0-x_1)^2}.$$ 
The term $f'(x_1)$ cannot disappear.
