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