Posted by **becca** on Tuesday, January 17, 2012 at 2:23am.

find the slope of the tangent line at the point P(1,1) on the graph of e^x-y=2x^2-y^2

- calculus -
**drwls**, Tuesday, January 17, 2012 at 6:00am
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

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