Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 149: 67

Answer

$6tan (sin^3t)\sec^2(sin^3t)\sin^2t \cos t $

Work Step by Step

y=$ tan^2(sin^3t) $ Apply the chain rule: $\frac{dy}{dt}=2tan(sin^3t).sec^2(sin^3t).(3sin^2t.(cost))$ =$6tan (sin^3t)\sec^2(sin^3t)\sin^2t \cos t $
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