Answer
\[{y^,} = 3\ln 7\,\left( {{7^{3x + 1}}} \right)\]
Work Step by Step
\[\begin{gathered}
y = {7^{3x + 1}} \hfill \\
Find\,\,the\,\,derivative \hfill \\
{y^,} = \,{\left( {{7^{3x + 1}}} \right)^,} \hfill \\
Use\,\,the\,\,formula \hfill \\
\frac{d}{{dx}}\,\,\left[ {{a^{g\,\left( x \right)}}} \right] = \,\left( {\ln a} \right){a^{g\,\left( x \right)}}{g^,}\,\left( x \right) \hfill \\
Then \hfill \\
Let\,\,a = 7\,,\,g\,\left( x \right) = 3x + 1 \hfill \\
{y^,} = \,\left( {\ln 7} \right){7^{3x + 1}}\,{\left( {3x + 1} \right)^,} \hfill \\
{y^,} = 3\ln 7\,\left( {{7^{3x + 1}}} \right) \hfill \\
\hfill \\
\end{gathered} \]
