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: 28

Answer

\[{p^,}\,\left( y \right) = \frac{1}{{y{e^y}}} - \frac{{\ln y}}{{{e^y}}}\]

Work Step by Step

\[\begin{gathered} p\,\left( y \right) = \frac{{\ln y}}{{{e^y}}} \hfill \\ \operatorname{Re} write\,\,the\,\,function \hfill \\ p\,\left( y \right) = {e^{ - y}}\ln y \hfill \\ Differentiate \hfill \\ {p^,}\,\left( y \right) = \,\,{\left[ {{e^{ - y}}\ln y} \right]^,} \hfill \\ Use\,\,the\,\,product\,\,rule \hfill \\ {p^,}\,\left( y \right) = {e^{ - y}}\,\,{\left[ {\ln y} \right]^,} + \ln y\,{\left( {{e^{ - y}}} \right)^,} \hfill \\ Then \hfill \\ {p^,}\,\left( y \right) = {e^{ - y}}\,\,\,\left( {\frac{1}{y}} \right) + \ln y\,\left( { - {e^{ - y}}} \right) \hfill \\ Multipliying \hfill \\ {p^,}\,\left( y \right) = \frac{{{e^{ - y}}}}{y} - {e^{ - y}}\ln y \hfill \\ {p^,}\,\left( y \right) = \frac{1}{{y{e^y}}} - \frac{{\ln y}}{{{e^y}}} \hfill \\ \end{gathered} \]
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