Answer
$y' = \frac{1}{2\sqrt x}secx + \sqrt {x} sec{x} tan{x}$
Work Step by Step
$y = \sqrt x secx +3$
$y' = \frac{1}{2\sqrt x}secx + \sqrt x sec{x}tanx$
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: 4
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