Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 147: 83


(a) $\dfrac{\pi}{360}$ (b) $1+\dfrac{\pi}{90}$

Work Step by Step

(a) Since \begin{align*} \left.\frac{d}{d \theta} \sin \theta\right|_{\theta=60^{\circ}}&=\left.\frac{d}{d \theta} \sin \left(\frac{\pi}{180} \theta\right)\right|_{\theta=60^{\circ}}\\ &=\left(\frac{\pi}{180}\right) \cos \left(\frac{\pi}{180}(60)\right)\\ &=\frac{\pi}{180} \frac{1}{2}\\ &=\frac{\pi}{360} \end{align*} (b) Since \begin{align*} \left.\frac{d}{d \theta}(\theta+\tan \theta)\right|_{\theta=45^{\circ}}&=\left.\frac{d}{d \theta}\left(\theta+\tan \left(\frac{\pi}{180} \theta\right)\right)\right|_{\theta=45^{\circ}}\\ &=1+\frac{\pi}{180} \sec ^{2}\left(\frac{\pi}{4}\right)\\ &=1+\frac{\pi}{90} \end{align*}
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