# 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^{y}dx/dx + xe^{y}dydx + e^{x}dydx + yd(e^{x })/dx = d(1)/dx

e^{y}(1) + xe^{y}dydx + e^{x}dydx + 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)

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