University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 344: 125


$-\dfrac{2e^{\cos (\ln x^2)}}{x}$

Work Step by Step

Here, $y'=-e^{\cos (\ln x^2)}(\dfrac{d (\ln x^2)}{dx}$ We need to apply the chain rule: Then $y'=-e^{\cos (\ln x^2)} (\dfrac{1}{x^2}) (2x)$ or, $y'=-\dfrac{2e^{\cos (\ln x^2)}}{x}$
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