## Calculus with Applications (10th Edition)

$y=14x-12$
$y^{4}(1-x)+xy= 2$ 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. $y-y_{2}=m(x-x_{1})$ $y-2=14(x-1)$ $y=14x-12$