## Calculus: Early Transcendentals (2nd Edition)

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