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: 16

Answer

\[{g^,}\,\left( z \right) = - \frac{1}{{{z^2} + 1}}\]

Work Step by Step

\[\begin{gathered} g\,\left( z \right) = {\tan ^{ - 1}}\,\left( {\frac{1}{z}} \right) \hfill \\ \hfill \\ Use\,\,the\,\,formula\,\,\,\,\frac{d}{{dz}}\,\,\left[ {{{\tan }^{ - 1}}u} \right] = \frac{{{u^,}}}{{1 + {u^2}}} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ {g^,}\,\left( z \right) = \frac{{ - \frac{1}{{{z^2}}}}}{{1 + \,{{\left( {\frac{1}{z}} \right)}^2}}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ {g^,}\,\left( z \right) = \frac{{ - \frac{1}{{{z^2}}}}}{{\frac{{{z^2} + 1}}{{{z^2}}}}} \hfill \\ \hfill \\ {g^,}\,\left( z \right) = - \frac{1}{{{z^2} + 1}} \hfill \\ \end{gathered} \]
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.