Answer
\[{y^,} = \frac{{3x + 3}}{{\,\left( {{x^2} + 2x} \right)\ln 3}}\]
Work Step by Step
\[\begin{gathered}
y = {\log _3}\,{\left( {{x^2} + 2x} \right)^{3/2}} \hfill \\
Use\,\,the\,\,property \hfill \\
\log {a^n} = n\log a \hfill \\
Then \hfill \\
y = \frac{3}{2}{\log _3}\,\left( {{x^2} + 2x} \right) \hfill \\
Find\,\,the\,\,derivative \hfill \\
{y^,} = \frac{3}{2}\,\,{\left[ {{{\log }_3}\,\left( {{x^2} + 2x} \right)} \right]^,} \hfill \\
Use\,\,the\,\,formula \hfill \\
\frac{d}{{dx}}\,\,\left[ {{{\log }_a}\left| {g\,\left( x \right)} \right|} \right] = \frac{1}{{\ln a}} \cdot \frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}} \hfill \\
Then \hfill \\
{y^,} = \frac{3}{2}\,\left( {\frac{1}{{\ln 3}}} \right)\,\left( {\frac{{\,{{\left( {{x^2} + 2x} \right)}^,}}}{{{x^2} + 2x}}} \right) \hfill \\
{y^,} = \frac{3}{{2\ln 3}}\,\left( {\frac{{2x + 2}}{{{x^2} + 2x}}} \right) \hfill \\
{y^,} = \frac{{3x + 3}}{{\,\left( {{x^2} + 2x} \right)\ln 3}} \hfill \\
\end{gathered} \]
