Answer
\[{y^,} = \frac{{6{x^3}\ln x - 2{x^3} + 2}}{{4x{{\ln }^2}x}}\]
Work Step by Step
\[\begin{gathered}
y = \frac{{{x^3} - 1}}{{2\ln x}} \hfill \\
Differentiate \hfill \\
{y^,} = \,\,{\left[ {\frac{{{x^3} - 1}}{{2\ln x}}} \right]^,} \hfill \\
Use\,\,the\,\,quotient\,\,rule \hfill \\
{y^,} = \frac{{\,\left( {2\ln x} \right)\,{{\left( {{x^3} - 1} \right)}^,} - \,\left( {{x^3} - 1} \right)\,\left( {2\ln x} \right)}}{{\,{{\left( {2\ln x} \right)}^2}}} \hfill \\
Then \hfill \\
{y^,} = \frac{{\,\left( {2\ln x} \right)\,\left( {3{x^2}} \right) - \,\left( {{x^3} - 1} \right)\,\left( 2 \right)\,\left( {\frac{1}{x}} \right)}}{{\,{{\left( {2\ln x} \right)}^2}}} \hfill \\
Simplifying \hfill \\
{y^,} = \frac{{6{x^2}\ln x - \frac{{2{x^3} - 2}}{x}}}{{4{{\ln }^2}x}} \hfill \\
{y^,} = \frac{{6{x^2}\ln x - \frac{{2{x^3} - 2}}{x}}}{{4{{\ln }^2}x}} \hfill \\
{y^,} = \frac{{6{x^3}\ln x - 2{x^3} + 2}}{{4x{{\ln }^2}x}} \hfill \\
\end{gathered} \]
