## Calculus: Early Transcendentals (2nd Edition)

$y = \frac{1}{3}x + \frac{5}{3}$
$\begin{gathered} \,{\left( {{x^2} + {y^2}} \right)^2} = \frac{{25}}{4}x{y^2}\,\,\,\,\,,\,\,\,\left( {1,2} \right) \hfill \\ \hfill \\ implicit\,\,differentiation \hfill \\ \hfill \\ 2\,\left( {{x^2} + {y^2}} \right)\,\left( {2x + 2y{y^,}} \right) = \frac{{25}}{2}xy{y^,} + \frac{{25}}{4}{y^2} \hfill \\ \hfill \\ multiply \hfill \\ \hfill \\ 4x\,\left( {{x^2} + {y^2}} \right) + 4y{y^,}\left( {{x^2} + {y^2}} \right) = \frac{{25}}{2}xy{y^,} + \frac{{25}}{4}{y^2} \hfill \\ \hfill \\ collect\,\,like\,\,terms \hfill \\ \hfill \\ 4y{y^,}\left( {{x^2} + {y^2}} \right) - \frac{{25}}{2}xy{y^,} = \frac{{25}}{4}{y^2} - 4x\,\left( {{x^2} + {y^2}} \right) \hfill \\ \hfill \\ solve\,\,for\,\,{y^,} \hfill \\ \hfill \\ {y^,} = \frac{{\frac{{25}}{4}{y^2} - 4x\,\left( {{x^2} + {y^2}} \right)}}{{4y\,\left( {{x^2} + {y^2}} \right) + \frac{{25}}{2}xy}} \hfill \\ \hfill \\ evaluate\,\,\,\left( {1,2} \right) \hfill \\ \hfill \\ then \hfill \\ \hfill \\ {y^,} = \frac{{\frac{{25}}{4}\,{{\left( 2 \right)}^2} - 4\,\left( 1 \right)\,\left( {{1^2} + {2^2}} \right)}}{{4\,\left( 2 \right)\,\left( {{1^2} + {2^2}} \right) - \frac{{25}}{2}\,\left( 1 \right)\,\left( 2 \right)}} = \frac{1}{3} \hfill \\ \hfill \\ use\,\,the\,\,point\, - \,slope\,\,form \hfill \\ \hfill \\ y - {y_1} = m\,\left( {x - {x_1}} \right) \hfill \\ \hfill \\ y - 2 = \frac{1}{3}\,\left( {x - 1} \right) \hfill \\ \hfill \\ y - 2 = \frac{1}{3}x - \frac{1}{3} \hfill \\ \hfill \\ y = \frac{1}{3}x + \frac{5}{3} \hfill \\ \hfill \\ \end{gathered}$