Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises - Page 191: 25


$y' = 3sec(3x+1)tan(3x+1)$

Work Step by Step

$y = sec (3x+1)$ $y' = (sec(3x+1))'$ The inner function is $g(x) = 3x+1$ and the outer function is $f(u) = sec (u)$ $f'(u) = sec(u)tan(u)$ $y' = (sec(3x+1))' = sec(g(x))tan(g(x)) g'(x)= sec(3x+1)tan(3x+1) \times 3= 3sec(3x+1)tan(3x+1)$
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.