Answer
$\frac{cosx}{\sqrt {1-sin^2x}} $
Work Step by Step
Let's apply the fundamental theorem of Calculus ($t=sinx$, and multiply by$(sin x)'=cosx)$. $\frac{dy}{dx}=(\frac{1}{\sqrt {1-sin^2x}})\cdot cosx=\frac{cosx}{\sqrt {1-sin^2x}} $
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 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 321: 51
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