University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 223: 38


a) $y=\tan \theta +C$ b) $y=\dfrac{2}{3} \sqrt{\theta^{3}}+C$ c) $y=\dfrac{2}{3} \sqrt{\theta^{3}}-\tan \theta +C$

Work Step by Step

a) When $\sec^2 \theta=(tan \theta)'$ Thus, $y'=\sec^2 \theta\implies y=\tan \theta +C$ b) When $\sqrt {\theta}=(\dfrac{2}{3} \sqrt{\theta^{3}})'$ Thus, $y'=\sqrt {\theta}\implies y=\dfrac{2}{3} \sqrt{\theta^{3}}+C$ c) The general solution will be the difference of part a and part b, that is, $y'=\dfrac{2}{3} \sqrt{\theta^{3}}-\tan \theta +C$
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