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: 62

Answer

$\frac{-5}{3}sin(5sin(\frac{t}{3}))(cos(\frac{t}{3}))$

Work Step by Step

y=$ cos(5sin(\frac{t}{3}))$ Apply the chain rule: $\frac{dy}{dt}=-sin(5sin(\frac{t}{3})).\frac{d}{dt}(5sin(\frac{t}{3}))$ =$-5(5sin(\frac{t}{3}))(5cos(\frac{t}{3})).\frac{d}{dt}\frac{t}{3}$ =$\frac{-5}{3}sin(5sin(\frac{t}{3}))(cos(\frac{t}{3}))$
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