Single Variable Calculus, Chapter 7, 7.3-2, Section 7.3-2, Problem 52

Determine the equation of the tangent line to the curve $\displaystyle xe^y + ye^x = 1$ at the point $(0,1)$

Solving for the slope


$
\begin{equation}
\begin{aligned}

& \frac{d}{dx} (xe^y) + \frac{d}{dx} (ye^x) = \frac{d}{dx} (1)
\\
\\
& \left[ (x) \frac{d}{dx} (e^y) + (e^y) \frac{d}{dx} (x) \right] + \left[ (y) \frac{d}{dx} (e^x) + (e^x) \frac{d}{dx} (y) \right] = 0
\\
\\
& x e^y \frac{dy}{dx} + e^y + ye^x + e^x \frac{dy}{dx} = 0
\\
\\
& xe^y y^1 + e^y + ye^x + e^x y^1 = 0
\\
\\
& xe^y y^1 + e^xy^1 = -e^y - ye^x
\\
\\
& y^1 (xe^y + e^x) = -e^y - ye^x
\\
\\
& y^1 = \frac{-e^y - ye^x}{xe^y + e^x}
\\
\\
& y^1 = \frac{-e^1 - 1 (e^0)}{0 (e^1) + e^0}
\\
\\
& y^1 = \frac{-e^1 - 1}{0 + 1}
\\
\\
& y^1 = -e - 1

\end{aligned}
\end{equation}
$



Using Point Slope Form


$
\begin{equation}
\begin{aligned}

y - y_1 =& m(x - x_1)
\\
\\
y - 1 =& (-e - 1) (x - 0)
\\
\\
y =& x (-e - 1) + 1
\\
\\
y =& 1 - x (e + 1)


\end{aligned}
\end{equation}
$