#### Answer

\[{y^,} = \frac{{2{x^3} - \sin 2x}}{{\sqrt {{x^4} + \cos 2x} }}\]

#### Work Step by Step

\[\begin{gathered}
y = \sqrt {{x^4} + \cos 2x} \hfill \\
\hfill \\
rewrite \hfill \\
\hfill \\
{y^,} = {\left( {{x^4} + \cos 2x} \right)^{1/2}} \hfill \\
\hfill \\
use\,\,the\,\,Chain\,\,rule \hfill \\
\hfill \\
{y^,} = \frac{1}{2}{\left( {{x^4} + \cos 2x} \right)^{ - 1/2}}{\left( {{x^4} + \cos 2x} \right)^,} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
{y^,} = \frac{{4{x^3} - \sin 2x \cdot \,{{\left( {2x} \right)}^,}}}{{2\sqrt {{x^4} + \cos 2x} }} \hfill \\
\hfill \\
{y^,} = \frac{{4{x^3} - 2\sin 2x}}{{2\sqrt {{x^4} + \cos 2x} }} \hfill \\
\hfill \\
{y^,} = \frac{{2{x^3} - \sin 2x}}{{\sqrt {{x^4} + \cos 2x} }} \hfill \\
\hfill \\
\end{gathered} \]