University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 25



Work Step by Step

$$s=\frac{4}{3\pi}\sin 3t+\frac{4}{5\pi}\cos5t$$ The derivative of function $s$ is: $$s'=\frac{4}{3\pi}(\sin 3t)'+\frac{4}{5\pi}(\cos5t)'=\frac{4}{3\pi}\cos 3t(3t)'+\frac{4}{5\pi}(-\sin5t)(5t)'$$ $$s'=\frac{4}{3\pi}\cos 3t\times3-\frac{4}{5\pi}\sin5t\times5=\frac{4}{\pi}\cos 3t-\frac{4}{\pi}\sin5t$$ $$s'=\frac{4}{\pi}(\cos3t-\sin5t)$$
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