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: 40

Answer

\[\frac{{dy}}{{dx}} = \frac{{3\pi {x^2}}}{{{x^3} + 1}}\]

Work Step by Step

\[\begin{gathered} y = \ln \,{\left( {{x^3} + 1} \right)^\pi } \hfill \\ \hfill \\ Use\,\,\,the\,\,property\,\,\ln {a^u} = u\ln a \hfill \\ \hfill \\ y = \pi \ln \,\left( {{x^3} + 1} \right) \hfill \\ \hfill \\ Use\,\,\frac{d}{{dx}}\,\,\left[ {\ln u} \right] = \frac{{{u^,}}}{u} \hfill \\ \hfill \\ y = \pi \frac{{{{\left( {{x^3} + 1} \right)}^,}\,}}{{{x^3} + 1}} \hfill \\ \hfill \\ Therefore \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \pi \,\left( {\frac{{3{x^2}}}{{{x^3} + 1}}} \right) \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{3\pi {x^2}}}{{{x^3} + 1}} \hfill \\ \end{gathered} \]
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