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: 27

Answer

\[{y^,} = \frac{{x{e^x}\ln x - {e^x}}}{{x{{\ln }^2}x}}\]

Work Step by Step

\[\begin{gathered} y = \frac{{{e^x}}}{{\ln x}} \hfill \\ Use\,\,the\,\,quotient\,\,rule \hfill \\ {y^,} = \frac{{\,\left( {\ln x} \right)\,{{\left( {{e^x}} \right)}^,} - {e^x}\,{{\left( {\ln x} \right)}^,}}}{{\,{{\left( {\ln x} \right)}^2}}} \hfill \\ Where\,\,\,{\left( {{e^x}} \right)^,} = {e^x}\,\,and\,\,\,{\left( {\ln x} \right)^,} = \frac{1}{x} \hfill \\ Then \hfill \\ {y^,} = \frac{{\,\left( {\ln x} \right)\,\left( {{e^x}} \right) - {e^x}\,\left( {1/x} \right)}}{{\,{{\left( {\ln x} \right)}^2}}} \hfill \\ Multipliying \hfill \\ {y^,} = \frac{{{e^x}\ln x - \frac{{{e^x}}}{x}}}{{{{\ln }^2}x}} \hfill \\ {y^,} = \frac{{x{e^x}\ln x - {e^x}}}{{x{{\ln }^2}x}} \hfill \\ \end{gathered} \]
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