Answer
\[{y^,} = \frac{{ - 2{x^2}}}{{2 - {x^2}}} + \ln \left| {2 - {x^2}} \right|\]
Work Step by Step
\[\begin{gathered}
y = x\ln \left| {2 - {x^2}} \right| \hfill \\
Differentiate \hfill \\
{y^,} = \,\,{\left[ {x\ln \left| {2 - {x^2}} \right|} \right]^,} \hfill \\
Use\,\,the\,\,product\,\,rule \hfill \\
{y^,} = x\,\,{\left[ {\ln \left| {2 - {x^2}} \right|} \right]^,} + \ln \left| {2 - {x^2}} \right|\,{\left( x \right)^,} \hfill \\
Use\,\,\frac{d}{{dx}}\,\,\left[ {\ln g\,\left( x \right)} \right] = \frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}} \hfill \\
{y^,} = x\,\left( {\frac{{\,{{\left( {2 - {x^2}} \right)}^,}}}{{2 - {x^2}}}} \right) + \ln \left| {2 - {x^2}} \right|\,\left( 1 \right) \hfill \\
{y^,} = x\,\left( {\frac{{\, - 2x}}{{2 - {x^2}}}} \right) + \ln \left| {2 - {x^2}} \right| \hfill \\
Multiply \hfill \\
{y^,} = \frac{{ - 2{x^2}}}{{2 - {x^2}}} + \ln \left| {2 - {x^2}} \right| \hfill \\
\end{gathered} \]
