University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.5 - Derivatives of Trigonometric Functions - Exercises - Page 151: 7


$y' = sinx +tanx\ secx=sinx(sec^2{x}+1)$

Work Step by Step

$y = sinx\ tanx$ $y' = cosx\ tanx + sinx\ sec^{2}x$ $y' = cosx\ \frac{sinx}{cosx} + sinx\ \frac{1}{cos^{2}x}$ $y' = sinx +tanx\ secx$
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