Calculus (3rd Edition)

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

Chapter 15 - Differentiation in Several Variables - 15.3 Partial Derivatives - Exercises - Page 781: 26


$$ A_\theta = 4\cos(4\theta-9t), A_t=-9\cos(4\theta-9t) . $$

Work Step by Step

Recall that $(\sin x)'=\cos x$. Since $ A=\sin(4\theta-9t)$, then by using the chain rule, we have $$ A_\theta = \cos(4\theta-9t) (4)=4\cos(4\theta-9t), \quad \\ A_t= cos(4\theta-9t) (-9)=-9\cos(4\theta-9t) . $$
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