Calculus: Early Transcendentals (2nd Edition)

$y = - \frac{1}{4}x + \frac{5}{4}$
$\begin{gathered} \sqrt[3]{x} + \sqrt[3]{{{y^4}}} = 2\,\,\,\,\,\,\,,\,\,\,\left( {1,1} \right) \hfill \\ \hfill \\ use\,\,the\,\,Implict\,\,\,differentiation \hfill \\ \hfill \\ \frac{1}{3}{x^{ - \frac{2}{3}}} + \frac{1}{3}\,{\left( {{y^4}} \right)^{ - \frac{2}{3}}}\,\left( {4{y^3}} \right){y^,} = 0 \hfill \\ \hfill \\ \frac{1}{{3{x^{\frac{2}{3}}}}} + \frac{4}{3}{y^{\frac{1}{3}}}{y^,} = 0 \hfill \\ \hfill \\ solve\,\,for\,\,{y^,} \hfill \\ \hfill \\ {y^,} = \,\left( {\frac{1}{{3{x^{\frac{2}{3}}}}}} \right)\,\left( {\frac{3}{{4{y^{\frac{1}{3}}}}}} \right) \hfill \\ \hfill \\ {y^,} = - \frac{1}{{4{x^{\frac{2}{3}}}{y^{\frac{1}{3}}}}} = - \frac{1}{4} \hfill \\ \hfill \\ use\,\,the\,\,po\operatorname{int} \,\,slope\,\,form \hfill \\ \hfill \\ y - {y_1} = m\,\left( {x - {x_1}} \right) \hfill \\ \hfill \\ y - 1 = - \frac{1}{4}\,\left( {x - 1} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ y = - \frac{1}{4}x + \frac{5}{4} \hfill \\ \end{gathered}$