Answer
$y' = 1 +tan^{2}x=sec^2{x}$
Work Step by Step
$y = (sinx +cosx)secx$
$y' = (cosx-sinx)secx + secx.tanx(sinx+cosx)$
$y' = cosx\ secx - sinx\ secx + secx\ tanx\ sinx + secx\ tanx\ cosx$
$y' = 1 - tanx +tan^{2}x + tanx$
$y' = 1 +tan^{2}x$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
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: 10
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