#### Answer

\[\frac{{dy}}{{dx}} = \frac{5}{{2y\,\left( {5 - {x^2}} \right)}}\]

#### Work Step by Step

\[\begin{gathered}
2{y^2} = \frac{{5 + x}}{{5 - x}} \hfill \\
\,Find\,\,dy/dx\,\,\,by\,\,implicit\,\,differentiation \hfill \\
\frac{d}{{dx}}\,\left( {2{y^2}} \right) = \frac{d}{{dx}}\,\left( {\frac{{5 + x}}{{5 - x}}} \right) \hfill \\
Use\,\,the\,\,power\,\,rule\,\,and\,\,quotient\,\,rule \hfill \\
4y{y^,} = \frac{{\,\left( {5 - x} \right)\,\left( 1 \right) - \,\left( {5 + x} \right)\,\left( { - 1} \right)}}{{\,{{\left( {5 - x} \right)}^2}}} \hfill \\
4y{y^,} = \frac{{5 - x + 5 + x}}{{\,{{\left( {5 - x} \right)}^2}}} \hfill \\
4y{y^,} = \frac{{10}}{{\,{{\left( {5 - x} \right)}^2}}} \hfill \\
{y^,} = \frac{{10}}{{4y\,{{\left( {5 - x} \right)}^2}}} \hfill \\
Then \hfill \\
\frac{{dy}}{{dx}} = \frac{5}{{2y\,\left( {5 - {x^2}} \right)}} \hfill \\
\hfill \\
\end{gathered} \]