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: 23

Answer

\[{f^,}\,\left( y \right) = - \frac{{2y}}{{{y^4} + 2{y^2} + 2}}\]

Work Step by Step

\[\begin{gathered} f\,\left( y \right) = {\cot ^{ - 1}}\,\left( {\frac{1}{{{y^2} + 1}}} \right) \hfill \\ \hfill \\ Use\,\,\frac{d}{{dy}}\,\,\,\left[ {{{\cot }^{ - 1}}\,u} \right] = \frac{{ - {u^,}}}{{1 + {u^2}}} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ {f^.}\,\left( y \right) = \frac{{ - \frac{{2y}}{{\,{{\left( {{y^2} + 1} \right)}^2}}}}}{{1 + \,{{\left( {\frac{1}{{{y^2} + 1}}} \right)}^2}}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ {f^,}\,\left( y \right) = \frac{{\frac{{ - 2y}}{{\,{{\left( {{y^2} + 1} \right)}^2}}}}}{{\frac{{\,{{\left( {{y^2} + 1} \right)}^2} + 1}}{{\,{{\left( {{y^2} + 1} \right)}^2}}}}} \hfill \\ \hfill \\ {f^,}\,\left( y \right) = - \frac{{2y}}{{{y^4} + 2{y^2} + 2}} \hfill \\ \hfill \\ \end{gathered} \]
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