Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.5 Derivatives of Trigonometric Functions - 3.5 Exercises: 38

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

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