Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.5 Linear Approximation and Differentials - 4.5 Exercises - Page 289: 45

Answer

$$dy = a\sin xdx$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = 2 - a\cos x,\,\,\,\,\,a{\text{ constant}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {2 - a\cos x} \right] \cr & f'\left( x \right) = 0 - a\frac{d}{{dx}}\left[ {\cos x} \right] \cr & f'\left( x \right) = - a\left( { - \sin x} \right) \cr & f'\left( x \right) = a\sin x \cr & {\text{Write in the form }}dy = f'\left( x \right)dx \cr & dy = a\sin xdx \cr} $$
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