Calculus 8th Edition

$y=1-3.718x$
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 $y=1-3.718x$