Answer
$y=-(e+1)(x+1)$
Work Step by Step
Find an equation of the tangent line to the curve $xe^{y}+ye^{x}=1$ at the point (0,1).
Differentiate $xe^{y}+ye^{x}=1$ with respect to x on both sides.
$xe^{y}\frac{dy}{dx}+e^{y}+ye^{x}+e^{x}\frac{dy}{dx}=0$
$y'=-\frac{e^{y}+ye^{x}}{e^{x}+xe^{x}}$
Let M be the slope of the equation at point (0,1).
$M=y'|_{(0,1)}=-(e+1)$
Therefore, the equation of the tangent at (0, 1) with slope$M=-(e+1)$ is given as:
$y=-(e+1)(x+1)$
