February 14th, 2009, 05:21 PM  #1 
Newbie Joined: Jul 2008 Posts: 6 Thanks: 0  Derivative Problem
Find y' when t equals 1 when y = (1 + r^2) ^ 1/2 and r = (t + 1)/(2t + 1) the answer in the book is  2/(9*13^1/2), but when I do it, using the power and chain rule, i get y' = 1/2(1 + r^2) ^ 1/2 * (1 + (( t + 1 )/(2t + 1))^2)' = 1/2(1 + r^2) ^ 1/2 * 2 * (( t + 1 ) / (2t + 1) ) / 2 then when I sub in r = (t + 1)/(2t + 1) and t = 4, i get (13/9)^1/2 * (1/3) = 1/(13^1/2) 
February 14th, 2009, 07:42 PM  #2 
Senior Member Joined: Dec 2008 Posts: 251 Thanks: 0  Re: Derivative Problem
The Chain Rule must also be applied to . I derived and When , and 
February 14th, 2009, 08:32 PM  #3 
Newbie Joined: Jul 2008 Posts: 6 Thanks: 0  Re: Derivative Problem
Wow. Awesome. Thanks a lot Scott !

February 16th, 2009, 04:18 PM  #4 
Senior Member Joined: Dec 2008 Posts: 251 Thanks: 0  Re: Derivative Problem
You're very welcome. 

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