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

Published by Pearson

# Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 27

#### Answer

$\frac{dr}{d\theta} = \frac{(cosec \theta)}{(cosec \theta +cot \theta)}$

#### Work Step by Step

$r = u^{-1}$ where, $u = (cosec \theta +cot \theta)$ Now, $\frac{dr}{du} = -u^{-2}$ and, $\frac{du}{d \theta} = -cosec \theta \times cot \theta - cosec ^{2} \theta = -cosec \theta(cot \theta + cosec \theta)$ So, $\frac{dr}{d \theta} = \frac{dr}{du} \times \frac{du}{d\theta}$ $\frac{dr}{d\theta} = -u^{-2}(-cosec \theta(cot \theta + cosec \theta))$ $\frac{dr}{d\theta} = (cosec \theta +cot \theta)^{-2}(cosec \theta(cot \theta + cosec \theta))$ $\frac{dr}{d\theta} = \frac{(cosec \theta(cot \theta + cosec \theta))}{(cosec \theta +cot \theta)^{2}}$ $\frac{dr}{d\theta} = \frac{(cosec \theta)}{(cosec \theta +cot \theta)}$

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