#### Answer

\[\frac{{dy}}{{dx}} = \,\left( {16\cos x - 8\sin x} \right)\,{\left( {\cos x + 2\sin x} \right)^7}\]

#### Work Step by Step

\[\begin{gathered}
y = \,{\left( {\cos x + 2\sin x} \right)^8} \hfill \\
\hfill \\
Use\,\,the\,\,version\,\,1\,\,of\,\,the\,\,chain\,\,rule \hfill \\
\hfill \\
{\text{ }}\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} \hfill \\
\hfill \\
set\,\,u = \cos x + 2\sin x \hfill \\
\,\,\,\,\,\,\,\,\frac{{du}}{{dx}} = - \sin x + 2\cos x \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = 8\,{\left( {\cos x + 2\sin x} \right)^7}\,\left( { - \sin x + 2\cos x} \right) \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \,\left( {16\cos x - 8\sin x} \right)\,{\left( {\cos x + 2\sin x} \right)^7} \hfill \\
\end{gathered} \]