## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 46

#### Answer

$$r'=\frac{\sec\sqrt\theta\tan\sqrt\theta\tan\frac{1}{\theta}}{2\sqrt\theta}-\frac{\sec\sqrt\theta\sec^2\frac{1}{\theta}}{\theta^2}$$

#### Work Step by Step

$$r=\sec\sqrt\theta\tan\frac{1}{\theta}$$ The derivative of function $r$ is $$r'=(\sec\sqrt\theta)'\tan\frac{1}{\theta}+\sec\sqrt\theta\Big(\tan\frac{1}{\theta}\Big)'$$ According to the Chain Rule, we have $$(\sec\sqrt\theta)'=\sec\sqrt\theta\tan\sqrt\theta(\sqrt\theta)'=\sec\sqrt\theta\tan\sqrt\theta(\theta^{1/2})'$$ $$=\frac{1}{2}\sec\sqrt\theta\tan\sqrt\theta(\theta^{-1/2})=\frac{\sec\sqrt\theta\tan\sqrt\theta}{2\sqrt\theta}$$ $$\Big(\tan\frac{1}{\theta}\Big)'=\sec^2\frac{1}{\theta}\Big(\frac{1}{\theta}\Big)'=\sec^2\frac{1}{\theta}\Big(\frac{-1(\theta)'}{\theta^2}\Big)=\sec^2\frac{1}{\theta}\Big(-\frac{1}{\theta^2}\Big)$$ $$=-\frac{1}{\theta^2}\sec^2\frac{1}{\theta}$$ Substitute these results back to the calculation of $r'$: $$r'=\frac{\sec\sqrt\theta\tan\sqrt\theta\tan\frac{1}{\theta}}{2\sqrt\theta}-\frac{\sec\sqrt\theta\sec^2\frac{1}{\theta}}{\theta^2}$$

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