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I am trying to improve my proving skills, started learning by myself, can anybody help me with this?

thanks you

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I'd just like to say, your question doesn't make it clear that it's related to minimum surfaces. –  SiliconCelery Oct 25 '13 at 18:21
    
I tried to re-tag this but ran into a snag with some "wiki edit priviledge" gizmo. But probably someone should! –  Robert Lewis Oct 25 '13 at 18:52
    
Please don't change your question to something else. If you have a new question, ask a new question. –  Javier Badia Oct 25 '13 at 19:04
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Also, what sort of course are you following that you can talk about minimal surfaces but don't know the derivative of $x^2+x$? –  Javier Badia Oct 25 '13 at 19:05
    
Why do you keep re-writing your questions? please stop! –  Fly by Night Oct 25 '13 at 23:27

2 Answers 2

One definition of a minimal surface is that its mean curvature is zero at each of its points. If $\kappa_1$ and $\kappa_2$ are the principal curvatures then $\tfrac{1}{2}(\kappa_1+\kappa_2)=0$.

Recall that the Gaussian curvature is given by the product $\kappa_1\kappa_2$.

If $\tfrac{1}{2}(\kappa_1+\kappa_2)=0$ then $\kappa_2 = -\kappa_1$ and so the Gaussian curvature is $\kappa_1\kappa_2 = -\kappa_1^2 \le 0$.

The case $\kappa_1=\kappa_2=0$ gives a plane and both the mean and the Gaussian curvatures are zero.

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can you show me proof that mean curative is zero ? –  user103234 Oct 25 '13 at 17:42
    
Minimal surfaces are defined to be surfaces with zero mean curvature: mathworld.wolfram.com/MinimalSurface.html See also: en.wikipedia.org/wiki/Minimal_surface#Definitions –  Fly by Night Oct 25 '13 at 17:44
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To Fly by Night: Did you perchance mean to type " . . . both the mean and the Gaussian curvatures are zero" in your final sentence? –  Robert Lewis Oct 25 '13 at 17:50
    
@RobertLewis Absolutely! Thanks for that. –  Fly by Night Oct 25 '13 at 17:52
    
@Fly by Night: Glad to help out; BUT, how the hell did this answer end up with this question? Any ideas? –  Robert Lewis Oct 25 '13 at 18:26

Disclaimer added at the advice of Javier Badia; see comments below: this answer was posted to a version of the question which was such a radical revision of the current (rolled back) question that it is really about a completely different topic. The question I answered was, basically: what is the derivative of $x^2 + x$? Apparently the OP put it through some rather severe revisions. So forgive me if this answer seems, well, rather curious for the current version of the question. Thanks for your patience. Robert Lewis

Oh well, while we're at it, might as well answer the new question, since I'm on this page and it's not too hard:

If

$y(x) = x^2 + x, \tag{1}$

then

$y'(x) = 2x + 1; \tag{2}$

and just for the sake of completeness:

given (1), we have

$y(x + h) = (x + h)^2 + (x + h); \tag{3}$

$y(x + h) = x^2 + 2xh + h^2 + x + h; \tag{4}$

$y(x + h) - y(x) = x^2 + 2xh + h^2 + x + h - x^2 - x; \tag{5}$

$y(x + h) - y(x) = 2xh + h^2 + h; \tag{6}$

$(y(x + h) - y(x)) / h = 2x + h + 1 = 2x + 1 + h; \tag{7}$

$y'(x) = \lim_{h \to 0}(y(x + h) - y(x)) / h = \lim_{h \to 0}(2x + h + 1) = 2x + 1; \tag{8}$

QED!!!

Hope this helps answer the new question! Cheerio,

and as always,

Fiat Lux!!!

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I rolled the question back to what it was originally. You should probably add a disclaimer to your answer so people don't start downvoting you. –  Javier Badia Oct 25 '13 at 19:08
    
@Javier Badia: Thanks, will do. –  Robert Lewis Oct 25 '13 at 19:15
    
@Javier Badia: how did you "roll back" the question? –  Robert Lewis Oct 25 '13 at 19:23
    
I just clicked "edit" and then "roll back" on the first revision, if that's what you're asking. –  Javier Badia Oct 25 '13 at 20:55
    
@Javier Badia: yeah, thanks, will give it a try! –  Robert Lewis Oct 25 '13 at 21:49

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