Calculus (3rd Edition)

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

Chapter 3 - Differentiation - Chapter Review Exercises - Page 163: 46


$$ y'= -\frac{4}{\theta^2}\cos\frac{4}{\theta}.$$

Work Step by Step

Recall that $(\sin x)'=\cos x$. Recall that $(x^n)'=nx^{n-1}$ Since $ y=\sin \frac{4}{\theta}$, then by the chain rule, the derivative $ y'$ is given by $$ y'= \cos\frac{4}{\theta} \left(-\frac{4}{\theta^2}\right)=-\frac{4}{\theta^2}\cos\frac{4}{\theta}.$$
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