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: 59

Answer

\[\frac{{dy}}{{dx}} = \frac{1}{{\sin 2x - 1}}\]

Work Step by Step

\[\begin{gathered} y = \frac{{\sin x}}{{\sin x - \cos x}} \hfill \\ \hfill \\ using\,\,quotient\,\,\,rule\,\,for\,\,derivatives \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{\cos x\,\left( {\sin x - \cos x} \right) - \sin x\,\left( {\cos x + \sin x} \right)}}{{\,{{\left( {\sin x - \cos x} \right)}^2}}} \hfill \\ \hfill \\ {\text{Therefore}}{\text{,}} \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{\cos x\sin x - {{\cos }^2}x - \sin x\cos x - {{\sin }^2}x}}{{{{\sin }^2}x - 2\sin x\cos x + {{\cos }^2}x}} \hfill \\ \hfill \\ {\text{Simplify}} \hfill \\ \hfill \\ use\,\,the\,\,identity\,\,:\,\,{\cos ^2}x + {\sin ^2}x = 1 \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{ - 1}}{{1 - 2\sin x\cos x}} = \frac{1}{{2\sin x\cos x - 1}} \hfill \\ \hfill \\ using\,:\,\,\sin 2x = 2\sin x\cos x \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{1}{{\sin 2x - 1}} \hfill \\ \hfill \\ \end{gathered} \]
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