Answer
\[{y^,} = \frac{{{x^2} - 2x}}{{\,{{\left( {x - 1} \right)}^2}}}\]
Work Step by Step
\[\begin{gathered}
y = \frac{{{x^2} - x + 1}}{{x - 1}} \hfill \\
Find\,\,the\,\,derivative\,\,of\,\,the\,\,function \hfill \\
{y^,} = \frac{d}{{dx}}\,\,\left[ {\frac{{{x^2} - x + 1}}{{x - 1}}} \right] \hfill \\
Use\,\,the\,\,quotient\,\,rule \hfill \\
{y^,} = \frac{{\,\left( {x - 1} \right)\,{{\left( {{x^2} - x + 1} \right)}^,} - \,\left( {{x^2} - x + 1} \right)\,{{\left( {x - 1} \right)}^,}}}{{\,{{\left( {x - 1} \right)}^2}}} \hfill \\
Then \hfill \\
{y^,} = \frac{{\,\left( {x - 1} \right)\,\left( {2x - 1} \right) - \,\left( {{x^2} - x + 1} \right)\,\left( 1 \right)}}{{\,{{\left( {x - 1} \right)}^2}}} \hfill \\
Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms\, \hfill \\
{y^,} = \frac{{2{x^2} - x - 2x + 1 - {x^2} + x - 1}}{{\,{{\left( {x - 1} \right)}^2}}} \hfill \\
{y^,} = \frac{{{x^2} - 2x}}{{\,{{\left( {x - 1} \right)}^2}}} \hfill \\
\end{gathered} \]
