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