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: 10

Answer

$y' = 1 +tan^{2}x=sec^2{x}$

Work Step by Step

$y = (sinx +cosx)secx$ $y' = (cosx-sinx)secx + secx.tanx(sinx+cosx)$ $y' = cosx\ secx - sinx\ secx + secx\ tanx\ sinx + secx\ tanx\ cosx$ $y' = 1 - tanx +tan^{2}x + tanx$ $y' = 1 +tan^{2}x$
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