Differentiability on the boundary 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?
 A: 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 $(0,+\infty)^n$.
A: 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 \, $.
