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: 36


a) $y=\sqrt x+C$ b) $y=2 \sqrt x+C$ c) $y=2x^2-2\sqrt x+C$

Work Step by Step

a) When $(\sqrt x)'=\dfrac{1}{2 \sqrt x}$ Thus, $y'=\dfrac{1}{2 \sqrt x}\implies y=\sqrt x+C$ b) When $( 2\sqrt x)'=\dfrac{1}{ \sqrt x}$ Thus, $y'=\dfrac{1}{\sqrt x}\implies y=2 \sqrt x+C$ c) When $( 2x^2-2\sqrt x)'=4x-\dfrac{1}{ \sqrt x}$ Thus, $y'=4x-\dfrac{1}{ \sqrt x} \implies y=2x^2-2\sqrt x+C$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.