calculus
posted by becca .
find the slope of the tangent line at the point P(1,1) on the graph of e^xy=2x^2y^2

What is the exponent of e: x or xy?
1,1 is a point on the curve of
e^(xy) = 2x^2 y^2
so I will use that.
Differentiate both sides implicitly with respect to x (treating y as an implicit function of x) and solve for the value of dy/dx when x = 1.
e^(xy)*(1  dy/dx) = 4x  2y*dy/dx
So at x = 1, y = 1,
1  dy/dx = 4  2 dy/dx
dy/dx = 3