University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 73



Work Step by Step

$$y=\frac{1}{9}\cot(3x-1)$$ - Find $y'$: $$y'=\frac{1}{9}\Big(\cot(3x-1)\Big)'=\frac{1}{9}\Big(-\csc^2(3x-1)\Big)(3x-1)'$$ $$y'=-\frac{1}{9}\csc^2(3x-1)\times3=-\frac{1}{3}\csc^2(3x-1)$$ - Find $y''$: $$y''=-\frac{1}{3}\Big(\csc^2(3x-1)\Big)'=-\frac{1}{3}\times2\csc(3x-1)\Big(\csc(3x-1)\Big)'$$ $$y''=-\frac{2}{3}\csc(3x-1)\Big(-\csc(3x-1)\cot(3x-1)(3x-1)'\Big)$$ $$y''=\frac{2}{3}\csc^2(3x-1)\cot(3x-1)\times3$$ $$y''=2\csc^2(3x-1)\cot(3x-1)$$
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.