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: 47

Answer

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

Work Step by Step

\[\begin{gathered} y = \sec x\csc x \hfill \\ \hfill \\ differentiate\,\,\,to\,\,find\,\,{y^{\,,}} \hfill \\ use\,\,the\,\,product\,\,rule \hfill \\ \hfill \\ y' = \csc x\,\left( {\sec x\tan x} \right) + \sec x\,\left( { - \csc x\cot x} \right) \hfill \\ \hfill \\ {\text{use}}\;\;{\text{trigonometric identities}} \hfill \\ \hfill \\ y' = \frac{1}{{\sin x}}\,\left( {\sec x\frac{{\sin x}}{{\cos x}}} \right) - \frac{1}{{\cos x}}\,\left( {\csc x\frac{{\cos x}}{{\sin x}}} \right) \hfill \\ \hfill \\ {\text{Simplify}} \hfill \\ \hfill \\ y' = {\sec ^2}x - {\csc ^2}x \hfill \\ \hfill \\ differentiate\,\,\,to\,\,find\,\,{y^{\,,,}} \hfill \\ \hfill \\ y'' = 2{\sec ^2}x\tan x + 2{\csc ^2}x\cot x \hfill \\ \hfill \\ y'' = 2\,\left( {{{\sec }^2}x\tan x + {{\csc }^2x}\cot x} \right) \hfill \\ \end{gathered} \]
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