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