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: 11

Answer

\[{f^,}\,\left( x \right) = - \frac{{2{e^{ - 2x}}}}{{\sqrt {1 - {e^{ - 4x}}} }}\]

Work Step by Step

\[\begin{gathered} f\,\left( x \right) = {\sin ^{ - 1}}\,\left( {{e^{ - 2x}}} \right) \hfill \\ \hfill \\ use\,\,the\,\,formula\,\,\frac{d}{{dx}}\,\,\left[ {{{\sin }^{ - 1}}\,u} \right] = \frac{{{u^,}}}{{\sqrt {1 - {u^2}} }} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ {f^,}\,\left( x \right) = \frac{{ - 2{e^{ - 2x}}}}{{\sqrt {1 - {e^{ - 4x}}} }} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ {f^,}\,\left( x \right) = - \frac{{2{e^{ - 2x}}}}{{\sqrt {1 - {e^{ - 4x}}} }} \hfill \\ \end{gathered} \]
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