Answer
\[\frac{{dy}}{{dx}} = 10x{\sec ^2}\,\left( {5{x^2}} \right)\]
Work Step by Step
\[\begin{gathered}
y = \tan \,\left( {5{x^2}} \right) \hfill \\
\hfill \\
Use\,\,the\,\,version\,\,1\,\,of\,\,the\,\,chain\,\,rule \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} \hfill \\
\hfill \\
set\,\,u = 5{x^2} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = {\sec ^2}\,\left( {5{x^2}} \right)\,\left( {10x} \right) \hfill \\
\hfill \\
multiply \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = 10x{\sec ^2}\,\left( {5{x^2}} \right) \hfill \\
\end{gathered} \]
