Let $x_0<x_1<...<x_n$, and let $f$ be continuously differentiable. Show that $$ \frac{\partial}{\partial_{x_i}} f[x_0,x_1,...,x_n]=f[x_0,x_1,...,x_i,x_i ,x_{i+1},...,x_n] $$.
I have the base case, namely
$n=1$: \begin{align*} \frac{\partial}{\partial_{x_0}}f[x_0,x_1] &=\frac{\partial}{\partial_{x_0}}\left[\frac{f(x_1)-f(x_0)}{x_1-x_0}\right]\\[1.5ex] &=\frac{f(x_1)-f(x_0)-f'(x_0)(x_1-x_0)}{(x_1-x_0)^2}\\[1.5ex] &=\frac{\frac{f(x_1)-f(x_0)}{x_1-x_0}-f'(x_0)}{x_1-x_0}\\[1.5ex] &=\frac{f[x_0,x_1]-f[x_0-x_0]}{x_1,x_0}\\[1.5ex] &=f[x_0,x_0,x_1] \end{align*} $n=k$:
Now assume that $$\frac{\partial}{\partial_{x_0}} f[x_0,x_1,...,x_k]=f[x_0,x_1,...,x_i,x_i ,x_{i+1},...,x_k]$$
$n=k+1$: $$\frac{\partial}{\partial_{x_0}} f[x_0,x_1,...,x_{k+1}]=f[x_0,x_1,...,x_i,x_i ,x_{i+1},...,x_{k+1}]$$
This is where I'm stuck. Thank you so much for any help.
The definition of the forward divided difference says that: $$f[x_0,x_1,\ldots,x_{n-1},x_n] \equiv \frac{f[x_1,\ldots,x_{n-1},x_n]-f[x_0,x_1,\ldots,x_{n-1}]}{x_{n}-x_0}$$
Now if $0<i<n$ then
$$\frac{\partial f[x_0,x_1,\ldots,x_{n-1},x_n]}{\partial x_i} = \frac{\frac{\partial f[x_1,\ldots,x_{n-1},x_n]}{\partial x_i}-\frac{\partial f[x_0,x_1,\ldots,x_{n-1}]}{\partial x_i}}{x_{n}-x_0}$$
By the induction hypotesis (for $n-1$) we get
$$\frac{\partial f[x_0,x_1,\ldots,x_{n-1},x_n]}{\partial x_i} = \frac{f[x_1,\ldots,x_i,x_i,x_{i+1},\ldots, x_n] - f[x_0,\ldots,x_i,x_i,x_{i+1},\ldots, x_{n-1}] }{x_{n}-x_0}$$
and by the definition of the forward divided difference the right hand side is
$$f[x_0,x_1,\ldots,x_i,x_i,x_{i+1},\ldots, x_n]$$
which is the same as the induction hypotesis (for $n$).