Answer
\[\frac{{dy}}{{dx}} = \frac{{5\sqrt {{x^4} + {y^2}} - 2{x^3}}}{{y - 6{y^2}\sqrt {{x^4} + {y^2}} }}\]
Work Step by Step
\[\begin{gathered}
\sqrt {{x^4} + {y^2}} = 5x + 2{y^3} \hfill \\
\hfill \\
implicit\,\,differentiation \hfill \\
\hfill \\
\frac{d}{{dx}}\,\,\left[ {\sqrt {{x^4} + {y^2}} } \right] = \frac{d}{{dx}}\,\,\left[ {5x + 2{y^3}} \right] \hfill \\
\hfill \\
then \hfill \\
\end{gathered} \]
\[\frac{{4{x^3} + 2yy'}}{{2\sqrt {{x^4} + {y^2}} }} = 5 + 6{y^2}y'\]
\[simplify\]
\[\frac{{2{x^3} + yy'}}{{\sqrt {{x^4} + {y^2}} }} = 5 + 6{y^2}y'\]
\[collect\,\,like\,\,terms\]
\[\frac{{yy'}}{{\sqrt {{x^4} + {y^2}} }} - 6{y^2}y' = 5 - \frac{{2{x^3}}}{{\sqrt {{x^4} + {y^2}} }}\]
\[factor\,\,{y^,}\]
\[\left( {\frac{y}{{\sqrt {{x^4} + {y^2}} }} - 6{y^2}} \right)y' = 5 - \frac{{2{x^3}}}{{\sqrt {{x^4} + {y^2}} }}\]
\[solve\,\,for\,\,{y^,}\]
\[\frac{{dy}}{{dx}} = \frac{{5 - \frac{{2{x^3}}}{{\sqrt {{x^4} + {y^2}} }}}}{{\frac{y}{{\sqrt {{x^4} + {y^2}} }} - 6{y^2}}}\]
\[\begin{gathered}
simplify \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{5\sqrt {{x^4} + {y^2}} - 2{x^3}}}{{y - 6{y^2}\sqrt {{x^4} + {y^2}} }} \hfill \\
\end{gathered} \]
