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 - Page 191: 32

Answer

\[\frac{{dy}}{{d\theta }} = - 4\sin \theta {\cos ^3}\theta + 4\cos \theta {\sin ^3}\theta \]

Work Step by Step

\[\begin{gathered} y = {\cos ^4}\theta + {\sin ^4}\theta \hfill \\ \hfill \\ Use\,\,the\,\,version\,\,2\,\,of\,\,the\,\,chain\,\,rule \hfill \\ \hfill \\ \,\,y = {u^n} \to \frac{{dy}}{{dx}} = n{u^{n - 1}}{u^,} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \frac{{dy}}{{d\theta }} = 4{\cos ^3}\,\left( \theta \right)\,\left( { - \sin \theta } \right) + 4{\sin ^3}\,\left( \theta \right)\,\left( {\cos \theta } \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{{dy}}{{d\theta }} = - 4\sin \theta {\cos ^3}\theta + 4\cos \theta {\sin ^3}\theta \hfill \\ \end{gathered} \]
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