Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.4 Exercises: 63

Answer

$y'=2\sec^2{(2x)}\cos{(\tan{2x})}.$

Work Step by Step

$u=\tan{2x}$; $\dfrac{du}{dx}=2\sec^2{2x}.$ $\dfrac{dy}{du}=\cos{u}$ $\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=(2\sec^2{2x})(\cos({\tan{2x})})$ $=2\sec^2{(2x)}\cos{(\tan{2x})}.$
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