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 Calculus Calculus Math Forum

 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.  Tags derivative, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post KMT Calculus 1 September 25th, 2011 06:55 PM azelio Algebra 2 October 17th, 2010 05:24 PM dagitt Calculus 3 January 5th, 2010 08:30 PM dagitt Calculus 1 January 2nd, 2010 04:17 PM StevenMx Calculus 6 February 4th, 2009 07:14 AM

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