Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - Review Exercises: 25

Answer

$\dfrac {\left( 2\sin x+2x\cos x\right) \left( 3x-1\right) +x\sin x}{\sqrt {3x-1}}=\dfrac {9x\sin x+6x^{2}\cos x-2\sin x-2x\cos x}{\sqrt {3x-1}} $

Work Step by Step

$\dfrac {d}{dx}\left( 2x\left( \sin x\right) \sqrt {3x-1}\right) =\left( \dfrac {d}{dx}2x\sin x\right) \times \sqrt {3x-1}+\left( \dfrac {d}{dx}\sqrt {3x-1}\right) \times 2x\sin x=\left( 2\sin x+2x\cos x\right) \times \sqrt {3x-1}+\dfrac {1}{2}\dfrac {3}{\sqrt {3x-1}}\times 2x\sin x= \dfrac {\left( 2\sin x+2x\cos x\right) \left( 3x-1\right) +3x\sin x}{\sqrt {3x-1}}=\dfrac {9x\sin x+6x^{2}\cos x-2\sin x-2x\cos x}{\sqrt {3x-1}} $
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.