Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.5 Derivatives of Trigonometric Functions - 3.5 Exercises - Page 169: 25

Answer

\[\frac{{dy}}{{dx}} = {\cos ^2}x - {\sin ^2}x\]

Work Step by Step

\[\begin{gathered} y = \sin x\cos x \hfill \\ \hfill \\ \,\,use\,\,the\,\,product\,\,rule. \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \,\left( {\sin x} \right)\,{\left( {\cos x} \right)^,} + \,\left( {\cos x} \right)\,{\left( {\sin x} \right)^,} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \,\left( {\sin x} \right)\,\left( { - \sin x} \right) + \,\left( {\cos x} \right)\,\left( {\cos x} \right) \hfill \\ \hfill \\ multiply \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = - {\sin ^2}x + {\cos ^2}x \hfill \\ \hfill \\ or \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = {\cos ^2}x - {\sin ^2}x \hfill \\ \end{gathered} \]
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