Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.10 Derivatives of Inverse Trigonometric Functions - 3.10 Execises: 25

Answer

\[{f^,}\,\left( x \right) = \frac{1}{{x\left| {\ln x} \right|\sqrt {{{\ln }^2}x - 1} }}\]

Work Step by Step

\[\begin{gathered} \hfill \\ f\,\left( x \right) = {\sec ^{ - 1}}\,\left( {\ln x} \right) \hfill \\ \hfill \\ Use\,\,\frac{d}{{dy}}\,\,\,\left[ {{{\sec }^{ - 1}}\,u} \right] = \frac{{{u^,}}}{{\left| u \right|\sqrt {{u^2} - 1} }} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ {f^,}\,\left( x \right) = \frac{{\,{{\left( {\ln x} \right)}^,}}}{{\left| {\ln x} \right|\sqrt {\,{{\left( {\ln x} \right)}^2} - 1} }} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ {f^,}\,\left( x \right) = \frac{{\frac{1}{x}}}{{\left| {\ln x} \right|\sqrt {{{\ln }^2}x - 1} }} \hfill \\ \hfill \\ {f^,}\,\left( x \right) = \frac{1}{{x\left| {\ln x} \right|\sqrt {{{\ln }^2}x - 1} }} \hfill \\ \end{gathered} \]
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