## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 3 - Derivatives - 3.5 Derivatives of Trigonometric Functions - 3.5 Exercises: 56

#### Answer

$\frac{{dy}}{{dx}} = \frac{1}{{1 + \cos x}}$

#### Work Step by Step

$\begin{gathered} y = \frac{{\sin x}}{{1 + \cos x}} \hfill \\ \hfill \\ differentiate\,\,use\,\,the\,\,quotient\,\,rule \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{\,\left( {1 + \cos x} \right)\,\left( {\cos x} \right) - \,\left( {\sin x} \right)\,\left( { - \sin x} \right)}}{{\,{{\left( {1 + \cos x} \right)}^2}}} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{\cos x + {{\cos }^2}x + {{\sin }^2}x}}{{\,{{\left( {1 + \cos x} \right)}^2}}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{\cos x + 1}}{{\,{{\left( {1 + \cos x} \right)}^2}}} \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{1}{{1 + \cos x}} \hfill \\ \hfill \\ \end{gathered}$

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