Answer
$y=14x-12$
Work Step by Step
$y^{4}(1-x)+xy= 2$
Find the y value: $y^{4}(1-1)+y= 2$
$y=2$
Calculate by implicit differentiation
$4y^{3}\frac{dy}{dx}-4xy^{3}\frac{dy}{dx}-y^{4}+x\frac{dy}{dx}+y=0$
$(4y^{3}-4xy^{3}+x)\frac{dy}{dx}=y^{4}-y$
$\frac{dy}{dx}=\frac{y^{4}-y}{4y^{3}-4xy^{3}+x}$
$m=14$
The equation of the tangent line is then found by using the point-slope form of the equation of a line.
$y-y_{2}=m(x-x_{1})$
$y-2=14(x-1)$
$y=14x-12$
