#### Answer

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

#### Work Step by Step

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

Published by
Pearson

ISBN 10:
0321999584

ISBN 13:
978-0-32199-958-0

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

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

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