#### Answer

\[\frac{{dy}}{{dx}} = \,\left( {12x + 3} \right)\cos \,\left( {4{x^3} + 3x + 1} \right)\]

#### Work Step by Step

\[\begin{gathered}
y = \sin \,\left( {4{x^3} + 3x + 1} \right) \hfill \\
\hfill \\
Use\,\,the\,\,version\,\,2\,\,of\,\,the\,\,chain\,\,rule\, \hfill \\
\hfill \\
\,\frac{d}{{dx}}\,\,\left[ {f\,\left( {g\,\left( x \right)} \right)} \right] = {f^,}\,\left( {g\,\left( x \right)} \right) \cdot {g^,}\,\left( x \right) \hfill \\
\hfill \\
set\,\,g\,\left( x \right) = 4{x^3} + 3x + 1 \hfill \\
\,\,\,\,\,\,\,\,\,\,\,{g^,}\,\left( x \right) = 12x + 3 \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \cos \,\left( {4{x^3} + 3x + 1} \right)\,\left( {12x + 3} \right) \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \,\left( {12x + 3} \right)\cos \,\left( {4{x^3} + 3x + 1} \right) \hfill \\
\end{gathered} \]