Answer
\[{y^,} = \frac{{\frac{{2{x^2}}}{{x + 3}} - 4x\ln \,\left( {x + 3} \right)}}{{{x^4}}}\]
Work Step by Step
\[\begin{gathered}
y = \frac{{2\ln \,\left( {x + 3} \right)}}{{{x^2}}} \hfill \\
Differentiate \hfill \\
{y^,} = \,\,\,\,{\left[ {\frac{{2\ln \,\left( {x + 3} \right)}}{{{x^2}}}} \right]^,} \hfill \\
Use\,\,the\,\,quotient\,\,rule \hfill \\
{y^,} = \frac{{{x^2}\,{{\left( {2\ln \,\left( {x + 3} \right)} \right)}^,} - 2\ln \,\left( {x + 3} \right)\,{{\left( {{x^2}} \right)}^,}}}{{\,{{\left( {{x^2}} \right)}^2}}} \hfill \\
Then \hfill \\
{y^,} = \frac{{{x^2}\,\left( {\frac{2}{{x + 3}}} \right) - 2\ln \,\left( {x + 3} \right)\,\left( {2x} \right)}}{{{x^4}}} \hfill \\
Simplify \hfill \\
{y^,} = \frac{{\frac{{2{x^2}}}{{x + 3}} - 4x\ln \,\left( {x + 3} \right)}}{{{x^4}}} \hfill \\
\end{gathered} \]
