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 - Page 169: 33

Answer

\[\frac{{dy}}{{dx}} = \sec x\tan x - \csc x\cot x\]

Work Step by Step

\[\begin{gathered} y = \sec x + \csc x \hfill \\ \hfill \\ differentiate \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{d}{{dx}}\,\,\left[ {\sec x} \right] + \frac{d}{{dx}}\,\,\left[ {\csc x} \right] \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \sec x\tan x - \csc x\cot x \hfill \\ \hfill \\ \end{gathered} \]
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