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: 26

Answer

The Derivative is: $\frac{dr}{d\theta}=\cos\theta+\sec^2\theta$

Work Step by Step

$r=(1+\sec\theta)\sin\theta$ Using the distributive property of mathematics $r=\sin\theta+\sec\theta\sin\theta$ $r=\sin\theta+\frac{1}{\cos\theta}\cdot\sin\theta$ $r=\sin\theta+\tan\theta$ Applying Derivative rules: $y'=f'(x)+g'(x)$ $\frac{dr}{d\theta}=\frac{d}{d\theta}(\sin\theta)+\frac{d}{d\theta}(\tan\theta)$ $\frac{dr}{d\theta}=(\cos\theta)+(\sec^2\theta)$ $\frac{dr}{d\theta}=\cos\theta+\sec^2\theta$
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