Answer
\[{k^,}\,\left( t \right) = \frac{{ - 7{x^2} - 14}}{{\,{{\left( {{x^2} - 2} \right)}^2}}}\]
Work Step by Step
\[\begin{gathered}
k\,\left( x \right) = \frac{{{x^2} + 7x - 2}}{{{x^2} - 2}} \hfill \\
Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,{k^,}\,\left( x \right) \hfill \\
{k^,}\,\left( x \right) = \frac{{\,\left( {{x^2} - 2} \right)\,{{\left( {{x^2} + 7x - 2} \right)}^,} - \,\left( {{x^2} + 7x - 2} \right)\,{{\left( {{x^2} - 2} \right)}^,}}}{{\,{{\left( {{x^2} - 2} \right)}^2}}} \hfill \\
Then \hfill \\
{k^,}\,\left( x \right) = \frac{{\left( {{x^2} - 2} \right)\,\left( {2x + 7} \right) - \left( {{x^2} + 7x - 2} \right)\,\,\left( {2x} \right)}}{{{{\left( {{x^2} - 2} \right)}^2}}} \hfill \\
Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\
{k^,}\,\left( x \right) = \frac{{2{x^3} + 7{x^2} - 4x - 14 - 2{x^3} - 14{x^2} + 4x}}{{\,{{\left( {{x^2} - 2} \right)}^2}}} \hfill \\
{k^,}\,\left( t \right) = \frac{{ - 7{x^2} - 14}}{{\,{{\left( {{x^2} - 2} \right)}^2}}} \hfill \\
\end{gathered} \]
