Answer
\[{y^,} = \frac{{ - 32{x^2} + 10x - 40}}{{\,{{\left( {4{x^2} - 5} \right)}^2}}}\]
Work Step by Step
\[\begin{gathered}
y = \frac{{ - {x^2} + 8x}}{{4{x^2} - 5}} \hfill \\
Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,{y^,} \hfill \\
{y^,} = \frac{{\,\left( {4{x^2} - 5} \right)\,{{\left( { - {x^2} + 8x} \right)}^,} - \,\left( { - {x^2} + 8x} \right)\,{{\left( {4{x^2} - 5} \right)}^,}}}{{\,{{\left( {4{x^2} - 5} \right)}^2}}}\, \hfill \\
Then, \hfill \\
{y^,} = \frac{{\,\,\,\left( {4{x^2} - 5} \right)\,\left( { - 2x + 8} \right) - \,\left( { - {x^2} + 8x} \right)\,\left( {8x} \right)}}{{\,{{\left( {4{x^2} - 5} \right)}^2}}} \hfill \\
Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\
{y^,} = \frac{{ - 8{x^3} + 32{x^2} + 10x - 40 + 8{x^3} - 64{x^2}}}{{\,{{\left( {4{x^2} - 5} \right)}^2}}} \hfill \\
{y^,} = \frac{{ - 32{x^2} + 10x - 40}}{{\,{{\left( {4{x^2} - 5} \right)}^2}}} \hfill \\
\end{gathered} \]
