#### Answer

\[{y^,} = \frac{{{x^2} - 2x - 1}}{{\,{{\left( {x - 1} \right)}^2}}}\]

#### Work Step by Step

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