Answer
\[{g^,}\,\left( z \right) = \,\left( {6{e^{2z}} + \frac{3}{z}} \right)\,{\left( {{e^{2z}} + \ln z} \right)^2}\]
Work Step by Step
\[\begin{gathered}
g\,\left( x \right) = \,{\left( {{e^{2z}} + \ln z} \right)^3} \hfill \\
Use\,\,the\,\,\,chain\,\,rule \hfill \\
{g^,}\,\left( z \right) = 3\,{\left( {{e^{2z}} + \ln z} \right)^{3 - 1}}\,{\left( {{e^{2z}} + \ln z} \right)^,} \hfill \\
Use\,\,the\,\,\,formula \hfill \\
\frac{d}{{dz}}\,\,{\left[ {{e^{g\,\left( z \right)}}} \right]^,} = {e^{g\,\left( z \right)}}{g^,}\,\left( z \right)\frac{d}{{dz}}\,\,\left[ {\ln z} \right] = \frac{1}{z} \hfill \\
Then \hfill \\
{g^,}\,\left( z \right) = 3\,{\left( {{e^{2z}} + \ln z} \right)^2}\,\left( {2{e^{2z}} + \frac{1}{z}} \right) \hfill \\
Multiplaying \hfill \\
{g^,}\,\left( z \right) = \,\left( {6{e^{2z}} + \frac{3}{z}} \right)\,{\left( {{e^{2z}} + \ln z} \right)^2} \hfill \\
\end{gathered} \]
