# Chapter 3 - Derivatives - 3.5 Derivatives of Trigonometric Functions - 3.5 Exercises: 61

$\frac{{dy}}{{dx}} = \frac{{2\sin x}}{{\,{{\left( {1 + \cos x} \right)}^2}}}$

#### Work Step by Step

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

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