Answer
\[{g^,}\,\left( z \right) = \,\,\frac{{3{z^2} + 1}}{{\,\left( {\ln 2} \right)\,\left( {{z^3} + z + 1} \right)}}\]
Work Step by Step
\[\begin{gathered}
g\,\left( z \right) = {\log _2}\,\left( {{z^3} + z + 1} \right) \hfill \\
Find\,\,the\,\,derivative\,\,of\,\,the\,\,\,function \hfill \\
{g^,}\,\left( z \right) = \,\,{\left[ {{{\log }_2}\,\left( {{z^3} + z + 1} \right)} \right]^,} \hfill \\
Use\,\,the\,\,formula \hfill \\
\frac{d}{{dx}}\,\,\left[ {{{\log }_a}g\,\left( x \right)} \right] = \frac{1}{{\ln a}}\,\left( {\frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}}} \right)\, \hfill \\
Then \hfill \\
{g^,}\,\left( z \right) = \,\,\,\left( {\frac{1}{{\ln 2}}} \right)\,\left( {\frac{{\,{{\left( {{z^3} + z + 1} \right)}^,}}}{{{z^3} + z + 1}}} \right) \hfill \\
{g^,}\,\left( z \right) = \,\,\left( {\frac{1}{{\ln 2}}} \right)\,\left( {\frac{{3{z^2} + 1}}{{{z^3} + z + 1}}} \right) \hfill \\
{g^,}\,\left( z \right) = \,\,\frac{{3{z^2} + 1}}{{\,\left( {\ln 2} \right)\,\left( {{z^3} + z + 1} \right)}} \hfill \\
\end{gathered} \]
