Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.9 Derivatives of Logarithmic and Exponential Functions - 3.9 Exercises: 15

Answer

\[ = - \frac{2}{{{x^2} - 1}}\]

Work Step by Step

\[\begin{gathered} \frac{d}{{dx}}\,\,\left[ {\ln \,\left( {\frac{{x + 1}}{{x - 1}}} \right)} \right] \hfill \\ \hfill \\ Use\,\,the\,\,\,property\,\,\ln \,\left( {\frac{a}{b}} \right) = \ln a - \ln b \hfill \\ \hfill \\ Therefore \hfill \\ \hfill \\ \frac{d}{{dx}} = \,\,\left[ {\ln \,\left( {x + 1} \right) - \ln \,\left( {x - 1} \right)} \right] \hfill \\ \hfill \\ Differentiate\,use\,\,\,the\,\,formula\,\,\frac{d}{{dx}}\,\,\left[ {\ln u} \right] = \frac{{{u^,}}}{u} \hfill \\ \hfill \\ = \frac{1}{{x + 1}} - \frac{1}{{x - 1}} \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ = \frac{{x - 1 - x - 1}}{{{x^2} - 1}} \hfill \\ \hfill \\ so \hfill \\ \hfill \\ = - \frac{2}{{{x^2} - 1}} \hfill \\ \hfill \\ R = \left\{ { - 1,1} \right\} \hfill \\ \hfill \\ \end{gathered} \]
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