## Calculus: Early Transcendentals (2nd Edition)

$y = \frac{2}{3}x + \frac{5}{3}$
$\begin{gathered} {x^4} - {x^2}y + {y^4} = 1\,\,\,\,\,\,\,\,,\,\,\,\left( { - 1,1} \right) \hfill \\ \hfill \\ use\,\,the\,\,implicit\,\,differentiation \hfill \\ \hfill \\ 4{x^3} - {x^2}y' - 2xy + 4{y^3}y' = 0 \hfill \\ \hfill \\ factor\,\,{y^,} \hfill \\ \hfill \\ y'\,\left( {4{y^3} - {x^2}} \right) = 2xy - 4{x^3} \hfill \\ \hfill \\ solve\,\,for\,\,y{\,^,} \hfill \\ \hfill \\ y' = \frac{{2xy - 4{x^3}}}{{4{y^3} - {x^2}}} \hfill \\ \hfill \\ evaluate\,\,the\,\,point\,\,\left( { - 1,1} \right) \hfill \\ \hfill \\ y' = \frac{{2\,\left( { - 1} \right)\,\left( 1 \right) - 4\,{{\left( { - 1} \right)}^3}}}{{4\,{{\left( 1 \right)}^3} - \,{{\left( { - 1} \right)}^2}}} \hfill \\ \hfill \\ use\,\,the\,\,point\, - \,slope\,\,form \hfill \\ \hfill \\ y - {y_1} = m\,\left( {x - {x_1}} \right) \hfill \\ \hfill \\ then \hfill \\ \hfill \\ y - 1 = \frac{{2\,\left( {x + 1} \right)}}{3} \hfill \\ \hfill \\ y = \frac{2}{3}x + \frac{5}{3} \hfill \\ \hfill \\ \end{gathered}$