My Math Forum derivative of extremum

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 May 12th, 2012, 01:02 AM #1 Newbie   Joined: Jan 2012 Posts: 3 Thanks: 0 derivative of extremum if the derivative dy/dx=0 at x=a then is it right to say that the derivative doesn't exist?
 May 12th, 2012, 03:23 AM #2 Member   Joined: Mar 2012 Posts: 60 Thanks: 0 Re: derivative of extremum That would be kind of delusive. The derivative is defined for that very value of x, but it just happen to equal zero in this case. I guess it is kind of arbitrary if you can make sure the reader know what you mean, as for many other mathematical wordings. If someone wrote that the derivative doesn't exist at a certain point I would suppose he/she meant that the derivative is not defined there. This may of course be due to my course litterature rather than my knowledge in english. The derivative can be plotted as a graph, and even though it intercepts the x-axis it actually "is there". I would rather say that the derivative vanishes at x=a. This word, on the other hand, implies that the derivative vanishes, and therefore do not exist at x=a, but this is the word I have seen being used for saying that the derivative is zero at a specific point or interval. I can't remember seeing any "official" words for how to describe that a derivative equals zero at some point, and I'm not sure there exist such regulations. To sum up, I would use other words than you, but I am not saying you are wrong if you can present your work to the reader without ambiguity. This is a subject that is very suiting to be open for discussion. It would be interesting to see other peoples preferences. Over and out.
 May 12th, 2012, 07:50 AM #3 Senior Member   Joined: Jun 2011 Posts: 298 Thanks: 0 Re: derivative of extremum Let $y=x+\frac{1}{x}$ $\frac{dy}{dx}=0\quad$ where $\quad x=\pm 2\quad$ and $\quad \frac{dy}{dx}$ doesn't exist at $x=0$ But where $y$ is a polynomial, $\frac{dy}{dx}$ exists on $(-\infty, \infty)$

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