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: 26


$y' = -csc(e^x)cot(e^x) \times e^x$

Work Step by Step

$y = csc (e^x)$ $y' = (csc (e^x))'$ The inner function is $g(x) = e^x$ and the outer function is $f(u) = csc(u)$ $f'(u) = -csc (u) cot ( u)$ $y' = (csc(e^x))' = -csc(g(x))cot(g(x)) g'(x)= -csc(e^x)cot(e^x) \times e^x$
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.