Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.5 Derivatives of Logarithmic Functions - 4.5 Exercises: 37

Answer

\[{y^,} = \frac{{3x + 3}}{{\,\left( {{x^2} + 2x} \right)\ln 3}}\]

Work Step by Step

\[\begin{gathered} y = {\log _3}\,{\left( {{x^2} + 2x} \right)^{3/2}} \hfill \\ Use\,\,the\,\,property \hfill \\ \log {a^n} = n\log a \hfill \\ Then \hfill \\ y = \frac{3}{2}{\log _3}\,\left( {{x^2} + 2x} \right) \hfill \\ Find\,\,the\,\,derivative \hfill \\ {y^,} = \frac{3}{2}\,\,{\left[ {{{\log }_3}\,\left( {{x^2} + 2x} \right)} \right]^,} \hfill \\ Use\,\,the\,\,formula \hfill \\ \frac{d}{{dx}}\,\,\left[ {{{\log }_a}\left| {g\,\left( x \right)} \right|} \right] = \frac{1}{{\ln a}} \cdot \frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}} \hfill \\ Then \hfill \\ {y^,} = \frac{3}{2}\,\left( {\frac{1}{{\ln 3}}} \right)\,\left( {\frac{{\,{{\left( {{x^2} + 2x} \right)}^,}}}{{{x^2} + 2x}}} \right) \hfill \\ {y^,} = \frac{3}{{2\ln 3}}\,\left( {\frac{{2x + 2}}{{{x^2} + 2x}}} \right) \hfill \\ {y^,} = \frac{{3x + 3}}{{\,\left( {{x^2} + 2x} \right)\ln 3}} \hfill \\ \end{gathered} \]
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