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April 1, 2015

April 1, 2015

Posted by **Jake** on Tuesday, February 5, 2013 at 12:51am.

g(t)=sqrt t(1+t)/t^2

- calc -
**Steve**, Tuesday, February 5, 2013 at 5:20amIs that √ everything, or

√(t(1+t)) / t^2, or

(√t)(1+t)/t^2 ?

- calc -
**Steve**, Tuesday, February 5, 2013 at 5:39amAh, I see from a prior post, we have

g(t) = √t (1+t)/t^2

you can expand that to be

g(t) = t^(-3/2) + t^(-1/2)

so the derivative is trivial:

g'(t) = -3/2 t^(-5/2) - 1/2 t^(-3/2)

= -(3+t) / (t^2 √t)

or you can use the product rule to get

g'(t) = 1/2√t (1+t)/t^2 + √t (1)/t^2 - 2√t (1+t)/t^3

= (1+t)/2t^2√t + 2t/2t^2√t - (4+4t)/2t^2√t

= (1+t+2t-4-4t)/2t^2√t

= -(3+t)/2t^2√t

- calc - find the typo -
**Steve**, Tuesday, February 5, 2013 at 5:40amfind the typo so the two solutions agree!

- calc -
**Jake**, Tuesday, February 5, 2013 at 9:16amagree thank you steve

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