T O P I C R E V I E W 
Kevitzinn 
Posted  10/19/2007 : 13:53:53 So I have, y = x(2x) and I find the derivative is
y' = (x)/(3(2x)^(2/3)) + (2x)
Is there some easy way to find the critical points of that? Thank you very much for your time. 
3 L A T E S T R E P L I E S (Newest First) 
Kevitzinn 
Posted  10/21/2007 : 23:53:39 Ohhhh, that's amazing! Thanks for the help, I would have never thought it could have been pulled out like that! Again, thanks a lot for the help, this site is the best. 
Kevitzinn 
Posted  10/19/2007 : 16:26:24 I see you have the right answer and I appreciate your time very much, but I have just one more question.
You said:
y' = (2  x)2/3[(x/3) + (2  x)]
What I'm confused about is, how did you do that? If x/3 is to the first power and and (2x) is to the (1/3) power, how can you pull it out like that? Thanks. 
skeeter 
Posted  10/19/2007 : 14:47:03 y = x(2  x)^{1/3}
y' = x*(1/3)(2  x)^{2/3}*(1) + (2  x)^{1/3}
y' = (2  x)^{2/3}[(x/3) + (2  x)]
y' = (2  x)^{2/3}[2  (4x/3)]
y' = [2  (4x/3)]/(2  x)^{2/3}
you should be able to almost "see" the critical values now.
