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February 14th, 2009, 05:21 PM   #1
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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)
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February 14th, 2009, 07:42 PM   #2
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Re: Derivative Problem

The Chain Rule must also be applied to . I derived



and



When ,



and

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February 14th, 2009, 08:32 PM   #3
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Re: Derivative Problem

Wow. Awesome. Thanks a lot Scott !
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February 16th, 2009, 04:18 PM   #4
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Re: Derivative Problem

You're very welcome.
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