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 - Page 321: 47


$-\frac{sinx}{2\sqrt x}$

Work Step by Step

Let's take the derivative of both sides, and let's apply the fundamental theorem of Calculus (substitute the limits, and multiply by their derivatives)! $\frac{dy}{dx}=sin(0^2)\cdot 0-sin(\sqrt x)^2\cdot(\frac{1}{2}x^{-\frac{1}{2}})=-\frac{sinx}{2\sqrt x}$
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