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: 19

Answer

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

Work Step by Step

\[\begin{gathered} f\,\left( x \right) = {\cos ^{ - 1}}\,\left( {\frac{1}{x}} \right) \hfill \\ \hfill \\ use\,\,\frac{d}{{dx}}\,\,\left[ {{{\cos }^{ - 1}}u} \right] = \frac{{ - {u^,}}}{{\sqrt {1 - {u^2}} }} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ {f^,}\,\left( x \right) = \frac{{ - \,\left( { - \frac{1}{{{x^2}}}} \right)}}{{\sqrt {1 - \,{{\left( {\frac{1}{x}} \right)}^2}} }} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ {f^,}\,\left( x \right) = \frac{{\frac{1}{{{x^2}}}}}{{\frac{{\sqrt {{x^2} - 1} }}{x}}} \hfill \\ \hfill \\ {f^,}\,\left( x \right) = \frac{1}{{x\sqrt {{x^2} - 1} }} \hfill \\ \end{gathered} \]
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