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I want to show that $$ \frac{d}{dx} f[x_0,\ldots, x_k, u_1,\ldots, u_n] = \sum_{v=1}^n f[x_0,\ldots, x_k, u_1,\ldots, u_n, u_v] \frac{d u_v}{d x} $$ where $x_0,\ldots x_k$ are constants and $u_1,\ldots, u_n$ are differentiable functions of $x$.

I understand that $$ f[x,x] = \lim_{\epsilon\rightarrow 0} f[x,x+\epsilon] = \lim_{\epsilon\rightarrow 0} \frac{f(x+\epsilon)-f(x)}{\epsilon} = f'(x) $$ but I'm not sure how to proceed when the divided difference has several arguments. In the following special case, would we write something like $$ f[x,x,x] = \lim_{\epsilon_1,\epsilon_2\rightarrow 0} f[x,x+\epsilon_1,x+\epsilon_2] = \lim_{\epsilon_1,\epsilon_2\rightarrow 0} \frac{f[x,x+\epsilon_2]-f[x,x+\epsilon_1]}{\epsilon_2-\epsilon_1}? $$

Alain
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  • I'm not sure I understand. Why the number of arguments changes in $f[]$? – Yuriy S Feb 17 '18 at 20:26
  • @YuriyS I think its saying that when you differentiate a divided difference, you add an extra argument. Alternatively, its a way to make sense of repeated arguments in divided differences. For example $f[x,x] = f'(x) = \frac{d}{dx} f[x]$. Just not sure how to extend the proof to the general case :( – Alain Feb 17 '18 at 23:32

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