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: 26

Answer

\[y' = \frac{{2x\sin x\,\left( {\sin x + 1} \right) + \,\left( {{x^2} - 1} \right)\cos x}}{{\,{{\left( {\sin x + 1} \right)}^2}}}\]

Work Step by Step

\[\begin{gathered} y = \frac{{\left( {{x^2} - 1} \right)\sin x}}{{\sin x + 1}} \hfill \\ \hfill \\ Using\,\,quotient\,\,rule \hfill \\ \hfill \\ y' = \frac{{\,\,{{\left[ {\,\left( {{x^2} - 1} \right)\sin x} \right]}^\prime } \cdot \,\,\left( {\sin x + 1} \right) - \,\left( {{x^2} - 1} \right)\sin x \cdot \,{{\left( {\sin x + 1} \right)}^\prime }}}{{\,{{\left( {\sin x + 1} \right)}^2}}} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ y' = \frac{{\,\,\left[ {2x\sin x + \,\left( {{x^2} - 1} \right)\cos x} \right]\,\left( {\sin x + 1} \right) - \,\left( {{x^2} - 1} \right)\sin x\cos x}}{{\,{{\left( {\sin x + 1} \right)}^2}}} \hfill \\ \hfill \\ y' = \frac{{2x\sin x\,\left( {\sin x + 1} \right) + \,\left( {{x^2} - 1} \right)\cos x\,\left( {\sin x + 1} \right) - \,\left( {{x^2} - 1} \right)\sin x\cos x}}{{\,{{\left( {\sin x + 1} \right)}^2}}} \hfill \\ \hfill \\ multiply \hfill \\ \hfill \\ y' = \frac{{2x\sin x\,\left( {\sin x + 1} \right) + \,\left( {{x^2} - 1} \right)\cos x}}{{\,{{\left( {\sin x + 1} \right)}^2}}} \hfill \\ \hfill \\ \end{gathered} \]
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