## Calculus: Early Transcendentals (2nd Edition)

$y = \frac{1}{2}x + 2$
$\begin{gathered} \,{\left( {x + y} \right)^{\frac{2}{3}}}\, = y\,\,\,,\,\,\,\,\left( {4,4} \right) \hfill \\ \hfill \\ use\,\,the\,\,Implicit\,\,differentiation \hfill \\ \hfill \\ \frac{2}{3}\,{\left( {x + y} \right)^{ - \frac{1}{3}}}\,\left( {1 + {y^,}} \right) = {y^,} \hfill \\ \hfill \\ distribute \hfill \\ \hfill \\ \frac{2}{{3\sqrt[3]{{x + y}}}} + \frac{{2{y^,}}}{{3\sqrt[3]{{x + y}}}} = {y^,} \hfill \\ \hfill \\ solve\,\,for\,\,{y^,} \hfill \\ \hfill \\ {y^,}\,\left( {\frac{2}{{3\sqrt[3]{{x + y}}}} - 1} \right) = - \frac{2}{{3\sqrt[3]{{x + y}}}} \hfill \\ \hfill \\ {y^,}\, = \frac{{\frac{2}{{3\sqrt[3]{{x + y}}}}}}{{\frac{2}{{3\sqrt[3]{{x + y}}}} - 1}} = - \frac{2}{{2 - 3\sqrt[3]{{x + y}}}} \hfill \\ \hfill \\ find\,\,the\,\,slope \hfill \\ \hfill \\ \,\left( {4,4} \right) \to \,{y^,} = - \frac{2}{{2 - 3\sqrt[3]{{4 + 4}}}} = \frac{1}{2} \hfill \\ \hfill \\ use\,\,the\,\,po\operatorname{int} \,\,slope\,\,form \hfill \\ \hfill \\ y - {y_1} = m\,\left( {x - {x_1}} \right) \hfill \\ \hfill \\ y - 4 = \frac{1}{2}\,\left( {x - 4} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ y = \frac{1}{2}x + 2 \hfill \\ \end{gathered}$