University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 49

Answer

$$y'=2\theta e^{-\theta^2}\sin(e^{-\theta^2})$$

Work Step by Step

$$y=\cos(e^{-\theta^2})$$ According to the Chain Rule, we have $$y=-\sin(e^{-\theta^2})(e^{-\theta^2})'=-\sin(e^{-\theta^2})(e^{-\theta^2})(-\theta^2)'$$ $$y'=-e^{-\theta^2}\sin(e^{-\theta^2})(-2\theta)$$ $$y'=2\theta e^{-\theta^2}\sin(e^{-\theta^2})$$
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