Answer
$y' = sinx +tanx\ secx=sinx(sec^2{x}+1)$
Work Step by Step
$y = sinx\ tanx$
$y' = cosx\ tanx + sinx\ sec^{2}x$
$y' = cosx\ \frac{sinx}{cosx} + sinx\ \frac{1}{cos^{2}x}$
$y' = sinx +tanx\ secx$
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: 7
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