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: 25

Answer

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

Work Step by Step

$r=\sec\theta\csc\theta$ Applying Derivative rules: $y'=f'(x)\cdot g(x)+f(x)\cdot g'(x)$ $\frac{dr}{d\theta}=(\csc\theta)\cdot\frac{d}{d\theta}(\sec\theta)+(\sec\theta)\cdot\frac{d}{d\theta}(\csc\theta)$ $\frac{dr}{d\theta}=(\csc\theta)(\sec\theta\tan\theta)+(\sec\theta)(-\csc\theta\cot\theta)$ $\frac{dr}{d\theta}=\sec\theta\csc\theta\tan\theta-\sec\theta\csc\theta\cot\theta$ $\frac{dr}{d\theta}=\sec\theta\csc\theta(\tan\theta-\cot\theta)$ Applying trigonometric identities $\frac{dr}{d\theta}=\frac{1}{\cos\theta}\frac{1}{\sin\theta}(\frac{sin\theta}{\cos\theta}-\frac{\cos\theta}{\sin\theta})$ $\frac{dr}{d\theta}=\frac{1}{\cos\theta\sin\theta}\cdot\frac{\sin\theta}{\cos\theta}-\frac{1}{\cos\theta\sin\theta}\cdot\frac{\cos\theta}{\sin\theta}$ $\frac{dr}{d\theta}=\frac{1}{\cos^2\theta}-\frac{1}{\sin^2\theta}$ $\frac{dr}{d\theta}=\sec^2\theta-\csc^2\theta$
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