Answer
\[\frac{{dy}}{{dx}} = \,\left( {200x + 220} \right)\,{\left( {5{x^2} + 11x} \right)^{19}}\]
Work Step by Step
\[\begin{gathered}
y = \,{\left( {5{x^2} + 11x} \right)^{20}} \hfill \\
\hfill \\
Use\,\,the\,\,version\,\,1\,\,of\,\,the\,\,chain\,\,rule \hfill \\
\hfill \\
{\text{ }}\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} \hfill \\
\hfill \\
set\,\,u = 5{x^2} + 11x \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = 20\,{\left( {5{x^2} + 11x} \right)^{19}}\frac{d}{{dx}}\,\,\left[ {5{x^2} + 11x} \right] \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = 20\,{\left( {5{x^2} + 11x} \right)^{19}}\,\left( {10x + 11} \right) \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \,\left( {200x + 220} \right)\,{\left( {5{x^2} + 11x} \right)^{19}} \hfill \\
\end{gathered} \]
