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: 46

Answer

\[y'' = 2{\sec ^2}x\tan x\]

Work Step by Step

\[\begin{gathered} y = \tan x \hfill \\ \hfill \\ find\,\,{y^{\,,}} \hfill \\ \hfill \\ y' = {\sec ^2}x \hfill \\ \hfill \\ differentiate\,\,to\,\,find\,\,{y^,}^, \hfill \\ use\,\,the\,\,chain\,\,\,rule \hfill \\ \hfill \\ y'' = 2\,\left( {\sec x} \right)\,{\left( {\sec x} \right)^\prime } \hfill \\ \hfill \\ so \hfill \\ \hfill \\ y'' = 2\,\left( {\sec x} \right)\,\left( {\sec x\tan x} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ y'' = 2{\sec ^2}x\tan x \hfill \\ \hfill \\ \end{gathered} \]
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