Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.10 Derivatives of Inverse Trigonometric Functions - 3.10 Execises - Page 221: 3


Slope of Tangent Line: $\frac{1}{5}$

Work Step by Step

$f(x) = \tan^{-1}(x)$ $f'(x) = \frac{1}{1+x^2}$ $f'(-2)=\frac{1}{1+(-2)^2} = \frac{1}{5}$
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.