Find an equation of the tangent line to the curve xe^y + ye^x = 1, (0, 1)

Solution:

xe^y + ye^x = 1

Differentiating the above equation w.r.t. x we have

e^ydx/dx + xe^ydydx + e^xdydx + yd(e^x )/dx = d(1)/dx

e^y(1) + xe^ydydx + e^xdydx + ye^x = 0

e^y + (xe^y + e^x)dydx + ye^x = 0

dy/dx = -( ye^x + e^y)/(xe^y + e^x)

The above equation is the tangent to the curve.

dy/dx at (0,1) = -(1.e⁰ + e¹)/(0.e¹ + e⁰) = -(1.1 + e)/(e) = -(1 + e)

Find an equation of the tangent line to the curve xe^y + ye^x = 1, (0, 1)

Summary:

The equation of the tangent line to the curve xe^y + ye^x = 1 is -( ye^x + e^y)/(xe^y + e^x). dy/dx at (0,1) = -(1 + e)