July 24th, 2012, 06:00 PM  #1 
Senior Member Joined: Jun 2011 Posts: 154 Thanks: 0  Derivative at extremum
Here is a problem I can't figure out how to work. I will post the problem in question, and then I will post the steps I completed so far. Between my notes, the video, and internet sources, I can't figure out what to do next Find the value of the derivative (if it exists) at the indicated extremum. (2/3 , 10(root3)/9) f(x) = 5x (rootx+1) fprime (2/3) = _________ Here's what I've done so far: (x^2+7)(2x)  (x^2)(2x) / (x^2+1)^2 2x^3+14x2x^3 / (x^2+7)^2 = 14x / (x^2+7)^2 (5x)(x+1) ^1/2 =(5x) * 1/2(x+1)^1/2 + (root x+1)(5) Pretty much don't know what to do next. I got this far by watching a video of a similar problem, but have no way to tell if I'm doing this right, since the video skips so many steps with no explanation as to how they got the values they're displaying. 
July 24th, 2012, 07:09 PM  #2 
Senior Member Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0  Re: Derivative at extremum
If I read that right, it's asking for the value of the derivative at a maximum or minimum value, and not at infinity or +infinity. In that case it will be zero no matter the function.

July 24th, 2012, 07:12 PM  #3 
Senior Member Joined: Jun 2011 Posts: 154 Thanks: 0  Re: Derivative at extremum
Thanks, I appreciate that. This class is a living nightmare lol. Somehow I'm still hanging on though.

July 24th, 2012, 07:49 PM  #4  
Senior Member Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0  Re: Derivative at extremum Quote:
 

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