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: 53

Answer

$y'=8\sec^2{x}\tan{x}.$

Work Step by Step

Using the Chain Rule: $u=\sec{x}$; $\dfrac{du}{dx}=\sec{x}\tan{x}$ $y=4u^2 ;\dfrac{dy}{du}=8u$ $\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=(8\sec{x})(\sec{x}\tan{x})=8\sec^2{x}\tan{x}.$
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