#### Answer

\[{p^,}\,\left( t \right) = \frac{{ - \frac{1}{{2\sqrt t }} - \frac{1}{2}\sqrt t }}{{\,{{\left( {t - 1} \right)}^2}}}\]

#### Work Step by Step

\[\begin{gathered}
p\,\left( t \right) = \frac{{\sqrt t }}{{t - 1}} \hfill \\
Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,{p^,}\,\left( t \right) \hfill \\
{p^,}\,\left( t \right) = \frac{{\,\left( {t - 1} \right)\,{{\left( {\sqrt t } \right)}^,} - \,\left( {\sqrt t \,} \right){{\left( {t - 1} \right)}^,}}}{{\,{{\left( {t - 1} \right)}^2}}} \hfill \\
Then \hfill \\
{p^,}\,\left( t \right) = \frac{{\,\left( {t - 1} \right)\,\left( {\frac{1}{{2\sqrt t }}} \right) - \sqrt t \,\left( 1 \right)}}{{\,{{\left( {t - 1} \right)}^2}}} \hfill \\
Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\
{p^,}\,\left( t \right) = \frac{{\frac{1}{2}{t^{1/2}} - \frac{1}{{2\sqrt t }} - \sqrt t }}{{\,{{\left( {t - 1} \right)}^2}}} \hfill \\
{p^,}\,\left( t \right) = \frac{{ - \frac{1}{{2\sqrt t }} - \frac{1}{2}\sqrt t }}{{\,{{\left( {t - 1} \right)}^2}}} \hfill \\
\end{gathered} \]