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: 34

Answer

\[ = \sec \,x\,\left( {{{\sec }^2}x + {{\tan }^2}x} \right)\]

Work Step by Step

\[\begin{gathered} y = \sec x\,\,\tan x \hfill \\ \hfill \\ Using\,\,product\,\,rule \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{d\sec x}}{{dx}}\tan x + \frac{{d\tan x}}{{dx}} \cdot \sec x \hfill \\ \hfill \\ then \hfill \\ \hfill \\ = \sec x \cdot \tan x \cdot \tan x + {\sec ^2}x \cdot \sec x \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \sec x \cdot {\tan ^2}x + {\sec ^3}x \hfill \\ \hfill \\ factor \hfill \\ \hfill \\ = \sec \,x\,\left( {{{\sec }^2}x + {{\tan }^2}x} \right) \hfill \\ \end{gathered} \]
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