Answer
$y=\frac{5}{2}x-\frac{1}{2}$
Work Step by Step
$y^{3}+xy-y= 8x^{4}$
Find the y value: $y^{3}+y-y= 8 $
$y=2$
Calculate by implicit differentiation
$3y^{2}\frac{dy}{dx}+x\frac{dy}{dx}+y -\frac{dy}{dx}=32x^{3}$
$(3y^{2}+x-1)\frac{dy}{dx}=32x^{3}-y$
$\frac{dy}{dx}=\frac{32x^{3}-y}{3y^{2}+x-1}$
$m=\frac{5}{2}$
The equation of the tangent line is then found by using the point-slope form of the equation of a line.
$y-y_{1}=m(x-x_{1})$
$y-2=\frac{5}{2}(x-1)$
$y=\frac{5}{2}x-\frac{1}{2}$
