Answer
$y=1-3.718x$
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).
$M=y'|_{(0,1)}=-(e+1)=-e-1=-3.718$
Therefore, the equation of the tangent at (0, 1) with slope
$M=-3.718$ is given as:
$y=-3.718x+1$
or $y=1-3.718x$