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: 55


$$\frac{dy}{dt}=10(t\tan t)^9\Big(\tan t+t\sec^2t\Big)$$

Work Step by Step

$$\frac{dy}{dt}=\frac{d}{dt}(t\tan t)^{10}$$ According to the Chain Rule: $$\frac{dy}{dt}=10(t\tan t)^9\frac{d}{dt}(t\tan t)=10(t\tan t)^9\Big(\tan t+t(\tan t)'\Big)$$ $$\frac{dy}{dt}=10(t\tan t)^9\Big(\tan t+t\sec^2t\Big)$$
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