Answer
\[{q^,} = 8{e^{2p + 1}}\,{\left( {{e^{2p + 1}} - 2} \right)^3}\]
Work Step by Step
\[\begin{gathered}
q = \,{\left( {{e^{2p + 1}} - 2} \right)^4} \hfill \\
Find\,\,the\,\,derivative\,\,of\,\,the\,\,\,function \hfill \\
q = \,\,{\left[ {{{\left( {{e^{2p + 1}} - 2} \right)}^4}} \right]^,} \hfill \\
Use\,\,the\,\,general\,\,\,power\,\,rule \hfill \\
{q^,} = 4{\left( {{e^{2p + 1}} - 2} \right)^3}{\left( {{e^{2p + 1}} - 2} \right)^,} \hfill \\
Use\,\,\frac{d}{{dx}}\,\,\left[ {{e^{g\,\left( x \right)}}} \right] = {e^{g\,\left( x \right)}}{g^,}\,\left( x \right) \hfill \\
Then \hfill \\
{q^,} = 4{\left( {{e^{2p + 1}} - 2} \right)^3}\,\left( {2{e^{2p + 1}}} \right) \hfill \\
Multiplying \hfill \\
{q^,} = 8{e^{2p + 1}}\,{\left( {{e^{2p + 1}} - 2} \right)^3} \hfill \\
\hfill \\
\hfill \\
\end{gathered} \]
