University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises: 51

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}} $
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