Answer
\[ = 2\theta \sec 5\theta + 5{\theta ^2}\sec 5\theta \tan 5\theta \]
Work Step by Step
\[\begin{gathered}
y = {\theta ^2}\sec 5\theta \hfill \\
\hfill \\
{y^,} = \frac{d}{{d\theta }}\,\left( {{\theta ^2}\sec 5\theta } \right) \hfill \\
\hfill \\
use\,\,the\,\,Product\,\,\,rule \hfill \\
\hfill \\
= \,{\left( {{\theta ^2}} \right)^,} \cdot \sec 5\theta + {\theta ^2} \cdot \,{\left( {\sec 5\theta } \right)^,} \hfill \\
\hfill \\
use\,\,the\,\,Chain\,\,rule \hfill \\
\hfill \\
= 2\theta \sec 5\theta + {\theta ^2} \cdot \,\left( {\sec 5\theta \tan 5\theta } \right) \cdot \,{\left( {5\theta } \right)^,} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= 2\theta \sec 5\theta + 5{\theta ^2}\sec 5\theta \tan 5\theta \hfill \\
\hfill \\
\end{gathered} \]
