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 141: 19


The Derivative is: $\frac{ds}{dt}=\sec^2 t-1$

Work Step by Step

$s=\tan t - t$ Applying derivative rules: $\frac{ds}{dt}=\frac{d}{dt}(\tan t)-\frac{d}{dt}(t)$ $\frac{ds}{dt}=(sec^2 t) -((1)t^{1-1})$ $\frac{ds}{dt}=\sec^2 t-1$
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.