Answer
\[{y^,} = \frac{{8x}}{{\ln 3\,\left( {{x^2} - 1} \right)}}\]
Work Step by Step
\[\begin{gathered}
y = 4\ln \left( 3 \right)\left( {{x^2} - 1} \right) \hfill \\
\hfill \\
Differentiate{\text{ }}Use\,\,\frac{d}{{dx}}\,\,\left[ {\ln u} \right] = \frac{{{u^,}}}{u} \hfill \\
\hfill \\
{y^,} = 4\,\left( {\frac{1}{{\ln \,\left( 3 \right)\,\left( {{x^2} - 1} \right)}}} \right)\,{\left( {{x^2} - 1} \right)^,} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
{y^,} = 4\,\left( {\frac{1}{{\ln 3\,\left( {{x^2} - 1} \right)}}} \right)\,\left( {2x} \right) \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
{y^,} = \frac{{8x}}{{\ln 3\,\left( {{x^2} - 1} \right)}} \hfill \\
\end{gathered} \]
