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: 20

Answer

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

Work Step by Step

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