Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises: 11

Answer

$\frac{2(sin(t) -sec^2(t))}{(cos(t)+tan(t))^3}$

Work Step by Step

Original Expression: $A(t) = \frac{1}{(cos(t) + tan(t))^2}$ $u = cos(t) + tan(t)$ $A(u) = \frac{1}{x^2}$ => $x^{-2}$ Apply the chain rule: A’(u)u’ $A’(u) = -2(u)^{-3}(-sin(t) + sec^2(t))$ $A’(t) = -2(cos(t) + tan(t))^{-3}(-sin(t) + sec^2(t))$ Simplify: $\frac{2(sin(t) -sec^2(t))}{(cos(t)+tan(t))^3}$
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.