Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Practice Exercises - Page 178: 59

Answer

$${\left. {\frac{{dw}}{{ds}}} \right|_{s = 0}} = \frac{{\sqrt 3 }}{8}$$

Work Step by Step

$$\eqalign{ & w = \sin \left( {\sqrt r - 2} \right){\text{ and }}r = 8\sin \left( {s + \pi /6} \right) \cr & {\text{Then}} \cr & w = \sin \left( {\sqrt {8\sin \left( {s + \pi /6} \right)} - 2} \right){\text{ }} \cr & {\text{find }}\frac{{dw}}{{ds}}{\text{ }} \cr & \frac{{dw}}{{ds}} = \frac{d}{{ds}}\left( {\sin \left( {\sqrt {8\sin \left( {s + \pi /6} \right)} - 2} \right){\text{ }}} \right) \cr & \frac{{dw}}{{ds}} = \cos \left( {\sin \left( {\sqrt {8\sin \left( {s + \pi /6} \right)} - 2} \right){\text{ }}} \right)\frac{d}{{ds}}\left( {\sqrt {8\sin \left( {s + \pi /6} \right)} - 2} \right) \cr & {\text{find the derivative}} \cr & \frac{{dw}}{{ds}} = \cos \left( {\sin \left( {\sqrt {8\sin \left( {s + \pi /6} \right)} - 2} \right){\text{ }}} \right)\left( {\frac{{\cos \left( {s + \pi /6} \right)}}{{2\sqrt {8\sin \left( {s + \pi /6} \right)} }}} \right) \cr & \cr & {\text{Evaluate at }}s = 0 \cr & {\left. {\frac{{dw}}{{ds}}} \right|_{s = 0}} = \cos \left( {\sin \left( {\sqrt {8\sin \left( {\pi /6} \right)} - 2} \right){\text{ }}} \right)\left( {\frac{{\cos \left( {\pi /6} \right)}}{{2\sqrt {8\sin \left( {\pi /6} \right)} }}} \right) \cr & {\left. {\frac{{dw}}{{ds}}} \right|_{s = 0}} = \cos \left( {\sin \left( {\sqrt 4 - 2} \right){\text{ }}} \right)\left( {\frac{{\sqrt 3 /2}}{{2\sqrt 4 }}} \right) \cr & {\left. {\frac{{dw}}{{ds}}} \right|_{s = 0}} = \cos \left( {\sin \left( 0 \right){\text{ }}} \right)\left( {\frac{{\sqrt 3 }}{8}} \right) \cr & {\left. {\frac{{dw}}{{ds}}} \right|_{s = 0}} = \frac{{\sqrt 3 }}{8} \cr} $$
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