posted by becca on .
find the slope of the tangent line at the point P(1,1) on the graph of e^x-y=2x^2-y^2
What is the exponent of e: x or x-y?
1,1 is a point on the curve of
e^(x-y) = 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^(x-y)*(1 - dy/dx) = 4x - 2y*dy/dx
So at x = 1, y = 1,
1 - dy/dx = 4 - 2 dy/dx
dy/dx = 3