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Let $U \subset \mathbb{R}^n$ open. If $f: U \to \mathbb{R}$ attains a relative maximum ( or minimum) in the point $x \in U$, and $f$ is differentiable in point $x$, then $f'(x)=0$.

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    I agree--I was being a bit $p$icky. It seems I like to see the right terminology. But I knew what you meant, and made a reply about the question in one of the "answers".2012-10-18

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This may be an approach: Take your function and create $n$ one variable functions by restricting $f$ to the coordinat axes near $x$. Apply the one variable result to each. Now you have all the partial derivatives zero; I seem to recall that "differentiable" in $n$ space implies the existence of a good tangent plane approximation to $f$, and from that and the zero partials, you might be able to get $f'(x)=0.$

If only I remembered my courses better.

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    I didn't put extra notation in my answer, but I meant that for the point x we use the lines that are *parallel* to the usual axes in n space, and pass through point x. To symbolize that would not be difficult, but it would take a lot of time/effort for me, not being much of an expert at LaTex.2012-10-18
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I assume that you know this result is valid for one dimensional functions. So you can proceed like this:

Consider the set $U_{d}=\{x+td:t\in (-\epsilon,\epsilon)\}$ where $\epsilon$ is a small number and $d\in\mathbb{R}^{n}$. Consider the restriction of $f$ to $U_{d}$; lets denote it by $f_{d}$.

Now, $f_{d}$ is a function of one variable and attains its maximum or minimum in $x$. Hence $f'_{d}(0)=0$, or equivalent $f'(x)d=0$. As $d$ is any point in $\mathbb{R}^{n}$, then you have $f'(x)=0$.

Ps: As you are brazilian, you can take a look in Elon's first book of analysis to see the proof for the case one dimensional.