Answer
\[\frac{{dy}}{{dw}} = \frac{{{{\sec }^2}w}}{{\,{{\left( {1 + \tan w} \right)}^2}}}\]
Work Step by Step
\[\begin{gathered}
y = \frac{{\tan w}}{{1 + \tan w}} \hfill \\
\hfill \\
{\text{using}}\,\,the\,\,quotient\,\,rule: \hfill \\
\hfill \\
\frac{{dy}}{{dw}} = \frac{{\left( {1 + \tan w} \right)\left[ {\tan w} \right]' - \left( {\tan w} \right)\left[ {1 + \tan w} \right]'}}{{{{\left( {1 + \tan w} \right)}^2}}} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\frac{{dy}}{{dw}} = \frac{{\,\left( {1 + \tan w} \right)\,\left( {{{\sec }^2}w} \right) - \tan w\,\left( {{{\sec }^2}w} \right)}}{{\,{{\left( {1 + \tan w} \right)}^2}}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
\frac{{dy}}{{dw}} = \frac{{{{\sec }^2}w + {{\sec }^2}w\tan w - {{\sec }^2}w\tan w}}{{\,{{\left( {1 + \tan w} \right)}^2}}} \hfill \\
\hfill \\
\frac{{dy}}{{dw}} = \frac{{{{\sec }^2}w}}{{\,{{\left( {1 + \tan w} \right)}^2}}} \hfill \\
\end{gathered} \]
