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: 58

Answer

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

Work Step by Step

\[\begin{gathered} y = \frac{1}{{2 + \sin x}} \hfill \\ \hfill \\ by\,\,the\,\,quotient\,\,rule \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{\left( {2 + \sin x} \right)\left[ 1 \right]' - \left( 1 \right)\left[ {2 + \sin x} \right]'}}{{{{\left( {2 + \sin x} \right)}^2}}} \hfill \\ then \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{0 - \,\left( {\cos x} \right)}}{{\,{{\left( {2 + \sin x} \right)}^2}}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = - \frac{{\cos x}}{{\,{{\left( {2 + \sin x} \right)}^2}}} \hfill \\ \end{gathered} \]
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