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: 41

Answer

\[ = 2\cos x - x\sin x\]

Work Step by Step

\[\begin{gathered} y = x\sin x \hfill \\ \hfill \\ differentiate\,\,to\,\,find\,\,{y^,} \hfill \\ use\,\,the\,\,product\,\,rule \hfill \\ \hfill \\ y' = x'\sin x + x\,{\left( {\sin x} \right)^\prime } \hfill \\ = \sin x + x\cos x \hfill \\ \hfill \\ differentiate\,\,to\,\,find\,\,{y^,}^, \hfill \\ use\,\,the\,\,product\,\,rule \hfill \\ \hfill \\ y'' = \cos x + x'\cos x + x\,{\left( {\cos x} \right)^\prime } \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ = \cos x + \cos x - x\sin x \hfill \\ \hfill \\ = 2\cos x - x\sin x \hfill \\ \hfill \\ \end{gathered} \]
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