Single Variable Calculus, Chapter 7, 7.2-1, Section 7.2-1, Problem 50

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

If we take the derivative of the curve implicitly, using Product Rule, we have


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$


Recall that the frist derivative is equal to the slope of the line at some point. So..

when $x = 0$ and $y = 1$,


$
\begin{equation}
\begin{aligned}

\frac{dy}{dx} =& \frac{-e^1 - (1) e^{(0)}}{0 (e^1) + e^{(0)}}
\\
\\
\frac{dy}{dx} =& -e^1 - 1

\end{aligned}
\end{equation}
$


Therefore, the equation of the tangent line can be determined using point slope form.


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$