Answer
\[{g^,}\,\left( x \right) = \frac{{4{x^2} + 2x - 12}}{{\,{{\left( {{x^2} + 3} \right)}^2}}}\]
Work Step by Step
\[\begin{gathered}
g\,\left( x \right) = \frac{{{x^2} - 4x + 2}}{{{x^2} + 3}} \hfill \\
Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,{g^,}\,\left( x \right) \hfill \\
{g^,}\,\left( x \right) = \frac{{\,\left( {{x^2} + 3} \right)\,{{\left( {{x^2} - 4x + 2} \right)}^,} - \,\left( {{x^2} - 4x + 2} \right)\,{{\left( {{x^2} + 3} \right)}^,}}}{{\,{{\left( {{x^2} + 3} \right)}^2}}} \hfill \\
Then, \hfill \\
{g^,}\,\left( x \right) = \frac{{\,\left( {{x^2} + 3} \right)\,\left( {2x - 4} \right) - \,\left( {{x^2} - 4x + 2} \right)\,\left( {2x} \right)}}{{\,{{\left( {{x^2} + 3} \right)}^2}}} \hfill \\
Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\
{g^,}\,\left( x \right) = \frac{{2{x^3} - 4{x^2} + 6x - 12 - 2{x^3} + 8{x^2} - 4x}}{{\,{{\left( {{x^2} + 3} \right)}^2}}} \hfill \\
{g^,}\,\left( x \right) = \frac{{4{x^2} + 2x - 12}}{{\,{{\left( {{x^2} + 3} \right)}^2}}} \hfill \\
\end{gathered} \]
