Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - Review Exercises: 22

Answer

$-\dfrac {36t\left( 3t^{2}+1\right) ^{2}}{\left( 3t^{2}-1\right) ^{4}}$

Work Step by Step

$\dfrac {d}{dt}\left( \dfrac {3t^{2}-1}{3t^{2}+1}\right) ^{-3}=-3\times \left( \dfrac {3t^{2}-1}{3t^{2}+1}\right) ^{-3-1}\times \dfrac {d}{dt}\left( \dfrac {3t^{2}-1}{3t^{2}+1}\right) =-3\left( \dfrac {3t^{2}-1}{3t^{2}+1}\right) ^{-4}\times \dfrac {6t\times \left( 3t^{2}+1\right) -6t\times \left( 3t^{2}-1\right) }{\left( 3t^{2}+1\right) ^{2}}=\dfrac {-36t}{\left( 3t^{2}+1\right) ^{2}}\left( \dfrac {3t^{2}-1}{3t+1}\right) ^{-4}=-\dfrac {36t\left( 3t^{2}+1\right) ^{2}}{\left( 3t^{2}-1\right) ^{4}}$
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