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: 33

Answer

\[\frac{{dy}}{{dx}} = 5\sec x\,{\left( {\sec x + \tan x} \right)^5}\]

Work Step by Step

\[\begin{gathered} y = \,\,{\left[ {\sec x + \tan x} \right]^5} \hfill \\ \hfill \\ Use\,\,the\,\,version\,\,2\,\,of\,\,the\,\,chain\,\,rule \hfill \\ \hfill \\ \,y = {u^n} \to \frac{{dy}}{{dx}} = n{u^{n - 1}}{u^,} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = 5\,{\left( {\sec x + \tan x} \right)^4}\,{\left( {\sec x + \tan x} \right)^,} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = 5\,{\left( {\sec x + \tan x} \right)^4}\,\left( {\sec x\tan x + {{\sec }^2}x} \right) \hfill \\ \hfill \\ factor \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = 5\sec x\,{\left( {\sec x + \tan x} \right)^5} \hfill \\ \end{gathered} \]
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.