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 - Page 158: 49


$$y'=-\frac{\sec t\tan t}{2\sqrt{1-\sec t}}$$ $$y''=-\frac{2\sec t(\tan^2t+\sec^2t)(1-\sec t)+\sec^2t\tan^2t}{4(1-\sec t)^{3/2}}$$

Work Step by Step

$$y'=\left(\sqrt{1-\sec t}\right)'=\frac{1}{2\sqrt{1-\sec t}}(1-\sec t)'= \frac{1}{2\sqrt{1-\sec t}}(1-\cos^{-1}t)'=\frac{1}{2\sqrt{1-\sec t}}(0-(-\cos^{-2}t)(\cos t)')=\frac{1}{2\sqrt{1-\sec t}}(\frac{1}{\cos^2t}\cdot(-\sin t))=-\frac{1}{2\sqrt{1-\sec t}}(\sec t\tan t)=-\frac{\sec t\tan t}{2\sqrt{1-\sec t}}$$ $$y''=(y')'=\left(-\frac{\sec t\tan t}{2\sqrt{1-\sec t}}\right)'=-\frac{(\sec t\tan t)'\cdot2\sqrt{1-\sec t}-\sec t\tan t\cdot(2\sqrt{1-\sec t})'}{(2\sqrt{1-\sec t})^2}$$ Let's first find derivatives $(\sec t\tan t)'$ and $(2\sqrt{1-\sec t})'$: $$(\sec t\tan t)'=(\sec t)'\tan t+\sec t(\tan t)'=\sec t\tan t\cdot \tan t+\sec t\cdot \sec^2t=\sec t(\tan^2 t+\sec^2t)$$ $$(2\sqrt{1-\sec t})'=2\frac{1}{2\sqrt{1-\sec t}}(1-\sec t)'=\frac{1}{\sqrt{1-\sec t}}(-\sec t\tan t)=-\frac{\sec t \tan t}{\sqrt{1-\sec t}}$$ Putting this into the derivative we get: $$y''=-\frac{\sec t(\tan^2t+\sec ^2t)\cdot2\sqrt{1-\sec t}-\sec t\tan t\cdot(-\frac{\sec t \tan t)}{\sqrt{1-\sec t}} }{4(1-\sec t)}= -\frac{2\sec t(\tan^2t+\sec^2t)(1-\sec t)+\sec^2t\tan^2t}{4(1-\sec t)^{3/2}}$$
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.