Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Review - Exercises: 11

Answer

$y'=\frac{cos\sqrt x}{2\sqrt x}-\frac{sin\sqrt x}{2}$

Work Step by Step

Start with the function: $y=\sqrt x cos\sqrt x$. Let $u=\sqrt x=x^{1/2}$. Then $y=u cos (u)$. Use the product rule to differentiate: $y'=u'cos(u)-uu'sin(u)$. Use power rule to find u': $u'=\frac{1}{2}x^{-1/2}$. Substitute expressions for u and u' into y' to find the answer: $y'=\frac{cos\sqrt x}{2\sqrt x}-\frac{sin\sqrt x}{2}$.
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