My Math Forum Derivative Problem

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 February 14th, 2009, 06:21 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, 08:42 PM #2 Senior Member   Joined: Dec 2008 Posts: 251 Thanks: 0 Re: Derivative Problem The Chain Rule must also be applied to $r\,=\,\frac{t\,+\,1}{2t\,+\,1}$. I derived $\begin{eqnarray*} \frac{dy}{dt} &=& \frac{d}{dt}\sqrt{1\,+\,r^2} \\ &=& \frac{1}{2\sqrt{1\,+\,r^2}}\cdot 2r\cdot \frac{dr}{dt} \\ &=& \frac{r}{\sqrt{1\,+\,r^2}} \cdot \frac{dr}{dt} \end{eqnarray*}$ and $\begin{eqnarray*} \frac{dr}{dt} &=& \frac{d}{dt}\left(\frac{t\,+\,1}{2t\,+\,1}\right) \\ &=& \frac{(2t\,+\,1)\cdot 1\,-\,(t\,+\,1)\cdot 2}{(2t\,+\,1)^2} \\ &=& \frac{2t\,+\,1\,-\,2t\,-\,2}{(2t\,+\,1)^2} \\ &=& \frac{-1}{(2t\,+\,1)^2}. \end{eqnarray*}$ When $t\,=\,1$, $r\,=\,\frac{1\,+\,1}{2\cdot1\,+\,1}\,=\,\frac{2}{3 }$ and $\frac{dr}{dt}\,=\,\frac{-1}{(2\cdot 1\,+\,1)^2}\,=\,-\frac{1}{3^2}\,=\,-\frac{1}{9}.$
 February 14th, 2009, 09:32 PM #3 Newbie   Joined: Jul 2008 Posts: 6 Thanks: 0 Re: Derivative Problem Wow. Awesome. Thanks a lot Scott !
 February 16th, 2009, 05:18 PM #4 Senior Member   Joined: Dec 2008 Posts: 251 Thanks: 0 Re: Derivative Problem You're very welcome.

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