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: 4

Answer

$y' = \frac{1}{2\sqrt x}secx + \sqrt {x} sec{x} tan{x}$

Work Step by Step

$y = \sqrt x secx +3$ $y' = \frac{1}{2\sqrt x}secx + \sqrt x sec{x}tanx$
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