Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises - Page 223: 10


$g'(t)=\frac{1}{2t\sqrt{1+\ln t}}$

Work Step by Step

$g(t)=\sqrt{1+\ln t}\\ g'(t)=\frac{1}{2\sqrt{1+\ln t}}\times\frac{1}{t}\\ g'(t)=\frac{1}{2t\sqrt{1+\ln t}}$
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.