Prove that $\frac{d}{dx}f[x_0,x_1,\ldots,x_{i-1},x]=f[x_0,x_1,\ldots,x_{i-1},x,x]$

Question

I have tried to prove it by induction on $i$-
For $i=1$, $f[x_0,x]=\frac{f(x)-f(x_0)}{x-x_0}=\frac{f(x)}{x-x_0}-\frac{f(x_0)}{x-x_0}$
So, $\frac{d}{dx}f[x_0,x]=\frac{f'(x)(x-x_0)-f(x)}{(x-x_0)^2}+\frac{f(x_0)}{(x-x_0)^2}$
$=\frac{f'(x)}{x-x_0}-\frac{f(x)}{(x-x_0)^2}+\frac{f(x_0)}{(x-x_0)^2}$ (as per definition $f[x,x]=f'(x)$)
$=\frac{f[x,x]}{x-x_0}-\frac{1}{x-x_0}\left(\frac{f(x)-f(x_0)}{x-x_0}\right)$
$=\frac{f[x,x]}{x-x_0}-\frac{f[x_0,x]}{x-x_0}$
$=-\frac{f[x_0,x]-f[x,x]}{x-x_0}=-f[x_0,x,x]$
But I'm getting a negetive sign before $f[x_0,x,x]$. Why it is happening? Am I doning any mistake? Can anyone solve the problem? Thanks for assistance in advance.

Answer 1

You get the error in the last line. Remember that $$ f[a,b,c]=\frac{f[a,b]-f[b,c]}{a-c} $$ then replace $a=b=x$ and $c=x_0$. Also remember that the order of points does not change the result.

You could get to this result much simpler by using $$ \frac{d}{dx}f[x_0,...,x_{i-1},x]=\lim_{z\to x}\frac{f[x_0,...,x_{i-1},z]-f[x_0,...,x_{i-1},x]}{z-x}=\lim_{z\to x}f[x_0,...,x_{i-1},z,x]. $$