## Calculus with Applications (10th Edition)

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