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: 18

Answer

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

Work Step by Step

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