#### Answer

\[{h^,}\,\left( z \right) = \frac{{11{z^{1.2}} - {z^{4.4}}}}{{{{\left( {{z^{3.2}} + 5} \right)}^2}}}\]

#### Work Step by Step

\[\begin{gathered}
h\,\left( z \right) = \frac{{{z^{2.2}}}}{{{z^{3.2}} + 5}} \hfill \\
Use\,\,the\,\,product\,\,rule\,\,to\,\,find\,\,{h^,}\,\left( z \right) \hfill \\
{h^,}\,\left( z \right) = \frac{{\,\left( {{z^{3.2}} + 5} \right)\,{{\left( {{z^{2.2}}} \right)}^,} - {z^{2.2}}{{\left( {{z^{3.2}} + 5} \right)}^,}\,}}{{\,{{\left( {{z^{3.2}} + 5} \right)}^2}}} \hfill \\
Then \hfill \\
{h^,}\,\left( z \right) = \frac{{\left( {{z^{3.2}} + 5} \right)\,\left( {2.2{z^{1.2}}} \right) - {z^{2.2}}\,\left( {3.2{z^{2.2}}} \right)}}{{{{\left( {{z^{3.2}} + 5} \right)}^2}}} \hfill \\
Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\
{h^,}\,\left( z \right) = \frac{{2.2{z^{4.4}} + 11{z^{1.2}} - {{3.2}^{4.4}}}}{{{{\left( {{z^{3.2}} + 5} \right)}^2}}} \hfill \\
{h^,}\,\left( z \right) = \frac{{11{z^{1.2}} - {z^{4.4}}}}{{{{\left( {{z^{3.2}} + 5} \right)}^2}}} \hfill \\
\hfill \\
\end{gathered} \]