Answer
\[\frac{{dy}}{{dt}} = \frac{{{{\sec }^2}t + {{\sec }^3}t - {{\tan }^2}t\sec t}}{{\,{{\left( {1 + \sec t} \right)}^2}}}\]
Work Step by Step
\[\begin{gathered}
y = \frac{{\tan t}}{{1 + \sec t}} \hfill \\
\hfill \\
{\text{using}}\,\,the\,\,quotient\,\,rule. \hfill \\
\hfill \\
\frac{{dy}}{{dt}} = \frac{{\left( {1 + \sec t} \right)\left[ {\tan t} \right]' - \left( {\tan t} \right)\left[ {1 + \sec t} \right]'}}{{{{\left( {1 + \sec t} \right)}^2}}} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\frac{{dy}}{{dt}} = \frac{{\,\left( {1 + \sec t} \right)\,\left( {{{\sec }^2}t} \right) - \tan t\,\left( {\sec t\tan t} \right)}}{{\,{{\left( {1 + \sec t} \right)}^2}}} \hfill \\
\hfill \\
multiply \hfill \\
\hfill \\
\frac{{dy}}{{dt}} = \frac{{{{\sec }^2}t + {{\sec }^3}t - {{\tan }^2}t\sec t}}{{\,{{\left( {1 + \sec t} \right)}^2}}} \hfill \\
\hfill \\
\end{gathered} \]
