University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.11 - Linearization and Differentials - Exercises - Page 199: 37

Answer

${dy}={\frac{-1}{|e^{(-x)}|{\sqrt{((e^{-x})^2-1})}}}\times{{(e^{(-x)})}}{dx}$

Work Step by Step

We are given: $y={\sec^{-1}{(e^{-x}})}$ on differentiating both sides: ${\frac{dy}{dx}}={\frac{d({\sec^{-1}{e^{-x}}})}{dx}}$ on applying the chain rule: ${\frac{dy}{dx}}={\frac{1}{|e^{(-x)}|{\sqrt{((e^{-x})^2-1})}}}\times{\frac{d(e^{(-x)})}{dx}}$ ${\frac{dy}{dx}}={\frac{1}{|e^{(-x)}|{\sqrt{((e^{-x})^2-1})}}}\times{{-(e^{(-x)})}}$ ${dy}={\frac{-1}{|e^{(-x)}|{\sqrt{((e^{-x})^2-1})}}}\times{{(e^{(-x)})}}{dx}$ The final answer is: ${dy}={\frac{-1}{|e^{(-x)}|{\sqrt{((e^{-x})^2-1})}}}\times{{(e^{(-x)})}}{dx}$
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.