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: 20

Answer

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

Work Step by Step

$s=t^2-\sec t+1$ Applying Derivative rules: $\frac{ds}{dt}=\frac{d}{dt}(t^2)-\frac{d}{dt}(\sec t)+\frac{d}{dt}(1)$ $\frac{ds}{dt}=(2)t^{2-1}-(\sec t\tan t)+(0)$ $\frac{ds}{dt}=2t-\sec t\tan t$
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