Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises: 42

Answer

\[\frac{{dy}}{{dx}} = \,\left( {16\cos x - 8\sin x} \right)\,{\left( {\cos x + 2\sin x} \right)^7}\]

Work Step by Step

\[\begin{gathered} y = \,{\left( {\cos x + 2\sin x} \right)^8} \hfill \\ \hfill \\ Use\,\,the\,\,version\,\,1\,\,of\,\,the\,\,chain\,\,rule \hfill \\ \hfill \\ {\text{ }}\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} \hfill \\ \hfill \\ set\,\,u = \cos x + 2\sin x \hfill \\ \,\,\,\,\,\,\,\,\frac{{du}}{{dx}} = - \sin x + 2\cos x \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = 8\,{\left( {\cos x + 2\sin x} \right)^7}\,\left( { - \sin x + 2\cos x} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \,\left( {16\cos x - 8\sin x} \right)\,{\left( {\cos x + 2\sin x} \right)^7} \hfill \\ \end{gathered} \]
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