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Suppose that we have a non negative real valued function $f$ defined only on $[0,\infty)^n$. Can one talk about the differentiability of such a function on the boundary? In the classical books on multivariable calculus, when they define differentiability of a multivariable function at a point, they always start with an assumption that the function is defined on an open neighborhood of the point. Can someone clear this for me?

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    the usual definition is: there exists an extension of $f$ to an open set containing $[0,\infty)^n$, which is as differentiable as you require2012-08-31
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    @user8268:Can $f$ be extended to an open set containing $[0,+\infty)^n$ such that their gradients agree on the boundary, i.e., if $\textbf{v}_0$ is a point on the boundary, $\nabla g(\textbf{v}_0)=\lim_{\textbf{v}\to \textbf{v}_0}\nabla f(\textbf{v})$?2012-08-31

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No you can't, at least it's not the usual differentiation.

It's like talking about the differentiation of $x \mapsto |x|$, for $x>0$ you can always define the derivative by:

$$ \lim_{h \to 0^+} \frac{f(x+h)-f(x)}{h}$$

In this case you get $1$. But if your map is defined in a neighborhood of $[0,+\infty)$: $(-\varepsilon, +\infty)$, for some $\varepsilon > 0$, this definition doesn't agree with the usual one.

In general, we just don't talk about differentiation in the boundary. It could be defined and continuous on $[0,+\infty)^n$ and differentiable on $(0,+\infty)^n$.

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    If I know that my $f$ is defined in $[0,+\infty)^n$ but differentiable in the interior $(0,+\infty)^n$, can I get an extension $g$ such that $g$ is differentiable in an open set containing $[0,+\infty)^n$ and $f=g$ on $[0,+\infty)^n$ and the gradient of $g$ at a boundary point say $(0.5,0,0,...,0)$ is equal to $\lim{\textbf{v}\to (0.5,0,0,...,0)}\nabla f(\textbf{v})$? (I am talking only about real valued functions here)2012-08-31
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    Yes of course. Let's see how it works in one variable. You define $g$ on $(-\varepsilon,0)$: for $x \in (-\varepsilon,0)$ to be $f(-x)$. Then $g$ has the properties you want.2012-08-31
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    Can you point some reference?2012-08-31
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    I don't know reference about this, as you said, usually we work on open sets. Nevertheless you can also do the construction using partition of unity, and you can easily find reference on it: mainly any book on differential geometry. http://en.wikipedia.org/wiki/Partition_of_unity2012-08-31
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I think the answer by Ilies Zidane above does not meet the OP question. There are two different settings here:

$(1)$ To extend the derivative $ \ f': \, ]0, \infty[ \to \mathbb{R} \ $ of a continuous function $ \ f: [0, \infty[ \, \to \mathbb{R} \ $ differentiable on the open interval $ \ ]0, \infty[ \ $ and defined only for nonnegative real numbers (hence nothing is assumed in the negative part of the real line) and

$(2)$ To talk about the derivative of a function $ \ g: \, ] \varepsilon , \infty [ \, \to \mathbb{R} \ $ (where $ \ \varepsilon > 0 \ $) whose restriction $ \ g|_{[0, \infty[} : [0, \infty[ \, \to \mathbb{R} \ $ is continuous and differentiable on the open interval $ \ ]0, \infty[ \ $.

In the first case we can extend the function $f$ in any way that suit us. In the second case there is an unique way we can extend $ \ g|_{[0, \infty[} \ $ in order to recover the already defined function $g \, $. As an example, in (1) we begin with $f$ defined as $ \ f(x)=x=|x| \ $ and we can extend it as $ \ x \mapsto x \ $ for nonnegative numbers. But in (2), if $ \ g(x)=|x| \ $, the same extension applied to $ \ g|_{[0, \infty[} \ $ does not recover $g \, $.