#### Answer

\[{y^,} = 8x - 20\]

#### Work Step by Step

\[\begin{gathered}
y = \,{\left( {2x - 5} \right)^2} \hfill \\
Write\,as\,\,a\,\,\,product \hfill \\
y = \,\left( {2x - 5} \right)\,\left( {2x - 5} \right) \hfill \\
Use\,\,the\,\,product\,\,rule\,\,to\,\,find\,\,{y^,} \hfill \\
{y^,} = \,\left( {2x - 5} \right)\,{\left( {2x - 5} \right)^,} + \,\left( {2x - 5} \right)\,{\left( {2x - 5} \right)^,} \hfill \\
{y^,} = 2\,\left( {2x - 5} \right)\,{\left( {2x - 5} \right)^,} \hfill \\
Then \hfill \\
{y^,} = 2\,\left( {2x - 5} \right)\,\left( 2 \right) \hfill \\
Multiplying \hfill \\
{y^,} = 4\,\left( {2x - 5} \right) \hfill \\
{y^,} = 8x - 20 \hfill \\
\end{gathered} \]