#### Answer

\[{f^,}\,\left( z \right) = \,\left( {4 - 4z{e^{ - {z^2}}}} \right)\,\left( {2z + {e^{ - {z^2}}}} \right)\]

#### Work Step by Step

\[\begin{gathered}
f\,\left( z \right) = \,{\left( {2z + {e^{ - {z^2}}}} \right)^2} \hfill \\
Use\,\,the\,\,generalized\,\,power\,\,rule \hfill \\
\frac{d}{{dx}}\,\,{\left[ {g\,\left( x \right)} \right]^n} = n\,\,{\left[ {g\,\left( x \right)} \right]^{n - 1}}{g^,}\,\left( x \right) \hfill \\
Then \hfill \\
{f^,}\,\left( z \right) = 2{\left( {2z + {e^{ - {z^2}}}} \right)^{2 - 1}}\,{\left( {2z + {e^{ - {z^2}}}} \right)^,} \hfill \\
Differentiate\,\,\,2z + {e^{ - {z^2}}} \hfill \\
{f^,}\,\left( z \right) = 2\left( {2z + {e^{ - {z^2}}}} \right)\,\left( {2 + {e^{ - {z^2}}}\,\left( { - 2z} \right)} \right) \hfill \\
Simplifying \hfill \\
{f^,}\,\left( z \right) = 2{\left( {2z + {e^{ - {z^2}}}} \right)^2}\,\left( {2 - 2z{e^{ - {z^2}}}} \right) \hfill \\
{f^,}\,\left( z \right) = \,\left( {4 - 4z{e^{ - {z^2}}}} \right)\,\left( {2z + {e^{ - {z^2}}}} \right) \hfill \\
\hfill \\
\end{gathered} \]