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: 78

Answer

\[{f^,}\,\left( x \right) = \frac{1}{x} - \frac{{6x}}{{{x^2} + 1}}\]

Work Step by Step

\[\begin{gathered} f\,\left( x \right) = \ln \frac{{2x}}{{\,{{\left( {{x^2} + 1} \right)}^3}}} \hfill \\ \hfill \\ Use\,\,\,the\,\,property\,\,\ln \,\left( {\frac{a}{b}} \right) = \ln a - \ln b \hfill \\ \hfill \\ then \hfill \\ \hfill \\ f\,\left( x \right) = \ln 2x - \ln \,{\left( {{x^2} + 1} \right)^3} \hfill \\ \hfill \\ f\,\left( x \right) = \ln 2x - 3\ln \,\left( {{x^2} + 1} \right) \hfill \\ \hfill \\ Differentiate{\text{ }}Use\,\,\frac{d}{{dx}}\,\,\left[ {\ln u} \right] = \frac{{{u^,}}}{u} \hfill \\ \hfill \\ {f^,}\,\left( x \right) = \frac{2}{{2x}} - 3\,\left( {\frac{{2x}}{{{x^2} + 1}}} \right) \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ {f^,}\,\left( x \right) = \frac{1}{x} - \frac{{6x}}{{{x^2} + 1}} \hfill \\ \end{gathered} \]
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