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: 18

Answer

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

Work Step by Step

\[\begin{gathered} \frac{d}{{dx}}\,\left( {\ln \,\left| {{x^2} - 1} \right|} \right) \hfill \\ \hfill \\ Use\,\,\frac{d}{{dx}}\,\,\left[ {\ln u} \right] = \frac{{{u^,}}}{u} \hfill \\ \hfill \\ set\,\,u = {x^2} - 1 \hfill \\ \hfill \\ Therefore \hfill \\ \hfill \\ = \frac{{2x}}{{{x^2} - 1}} \hfill \\ \end{gathered} \]
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