# Chapter 4 - Calculating the Derivative - 4.2 Derivatives of Products and Quotients - 4.2 Exercises - Page 216: 19

${g^,}\,\left( x \right) = \frac{{4{x^2} + 2x - 12}}{{\,{{\left( {{x^2} + 3} \right)}^2}}}$

#### Work Step by Step

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

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