Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Review - Exercises: 31

Answer

$\dfrac {4x}{1+16x^{2}}+\tan ^{-1}4x$

Work Step by Step

$\dfrac {d}{dx}\left( x\tan ^{-1}\left( 4x\right) \right) =\left( \dfrac {d}{dx}\left( x\right) \right) x\tan ^{-1}\left( 4x\right) +xx\left( \dfrac {d}{dx}\tan ^{-1}\left( 4x\right) \right) =\tan ^{-1}4x+\dfrac {x}{1+\left( 4x\right) ^{2}}\times \dfrac {d}{dx}\left( 4x\right) =\dfrac {4x}{1+16x^{2}}+\tan ^{-1}4x$
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