Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Review - Exercises: 19

Answer

$\dfrac {1-t^{2}}{\left( 1+t^{2}\right) ^{2}}sec^{2}\dfrac {t}{1+t^{2}}$

Work Step by Step

$\dfrac {d}{dt}\tan \left( \dfrac {t}{1+t^{2}}\right) =\dfrac {1}{\cos ^{2}\left( \dfrac {t}{1+t^{2}}\right) }\times \dfrac {d}{dt}\left( \dfrac {t}{1+t^{2}}\right) =sec^2\dfrac {t}{1+t^{2}}\times \dfrac {\left( \dfrac {d}{dt}\left( t\right) \right) \times \left( 1+t^{2}\right) -\dfrac {d}{dt}\left( 1+t^{2}\right) \times t}{\left( 1+t^{2}\right) ^{2}}= \dfrac {1-t^{2}}{\left( 1+t^{2}\right) ^{2}}sec^{2}\dfrac {t}{1+t^{2}}$
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