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