Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.5 - Derivatives of Trigonometric Functions - Exercises 3.5 - Page 142: 44

Answer

$y=-x+\frac{\pi}{2}$ See graph.

Work Step by Step

Step 1. Given $y=cot(x)$, we have $y'=-csc^2x$ Step 2. For a tangent line parallel to the line $y=-x$, let $y'=-1$, we have $-csc^2x=-1$ and $csc(x)=\pm1$ thus $x=\pi/2$ in $(0,\pi)$ Step 3. At $x=\pi/2$, $y=cot(\pi/2)=0$, and the tangent line equation is $y=-(x-\pi/2)$ or $y=-x+\frac{\pi}{2}$ Step 4. See graph.
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.