how do go about finding the slope of the tangent to x^2y^2 + xy = 2 at (1,1)?

Question

thanks

multiple choice answers are 2.01, -2.01,1,-1,0 or 3.14

Answer 1

Differentiate, implicitly, using the product rule a few times.
x^2 y^2 differentiates to (2x)*(y^2) + (x^2)*(2y * y').
xy differentiates to (1)*(y) + (x)*(y').
2 differentiates to 0.
So 2xy^2 + 2x^2yy' + y + xy' = 0.
Substitute in x=1, y=1:
2 + 2y' + 1 + y' = 0
And solve for y':
3y' = -3
y' = -1.

Answer 2

x^2*2yy'+2xy^2+xy'+y=0
at (1,1)
2y'+2+y'+1=0
3y'=-3
y'=-1
slope=-1