Find y' for xe^(-y/2)+ye^(-x/2)-2=0, and evaluate y'(0)

Find $y'$ for $xe^{-y/2}+ye^{-x/2}-2=0$, and evaluate $y'(0)$.

Step 1: Find $y$ at $x = 0$

Setting $x = 0$ kills the first term and reduces $e^{-x/2}$ to $1$: $0 + y\cdot 1 - 2 = 0 \;\Longrightarrow\; y = 2$.

So the point is $(0,2)$.

Step 2: Differentiate the first term

Product rule, with $\bigl(e^{-y/2}\bigr)' = -\tfrac{y'}{2}e^{-y/2}$: \[\frac{d}{dx}\bigl(xe^{-y/2}\bigr) = e^{-y/2} + x\cdot\left(-\frac{y'}{2}\right)e^{-y/2} = e^{-y/2}\left(1-\frac{xy'}{2}\right).\]

Step 3: Differentiate the second term

Product rule again, now with $\bigl(e^{-x/2}\bigr)' = -\tfrac12 e^{-x/2}$: \[\frac{d}{dx}\bigl(ye^{-x/2}\bigr) = y'e^{-x/2} - \frac{y}{2}e^{-x/2} = e^{-x/2}\left(y'-\frac{y}{2}\right).\]

Step 4: Assemble and solve

\[ e^{-y/2}\left(1-\frac{xy'}{2}\right) + e^{-x/2}\left(y'-\frac{y}{2}\right) = 0. \]

Collecting the $y'$ terms on one side: $y'\left(e^{-x/2}-\frac{x}{2}e^{-y/2}\right) = \frac{y}{2}e^{-x/2} - e^{-y/2}$.

\[ y' = \frac{\dfrac{y}{2}e^{-x/2}-e^{-y/2}}{e^{-x/2}-\dfrac{x}{2}e^{-y/2}}. \]

Step 5: Evaluate at $(0,2)$

Here $e^{-x/2} = 1$ and $e^{-y/2} = e^{-1}$, and the $x$-term in the denominator drops out: $y'(0) = \frac{1\cdot 1 - e^{-1}}{1-0} = 1-\frac1e \approx 0.63212$.

Step 6: Verification

\[x = 0.001:\quad \text{solving } xe^{-y/2}+ye^{-x/2} = 2 \text{ gives } y \approx 2.000632\]

\[\frac{2.000632-2}{0.001} \approx 0.632 \approx 1-\frac1e \;\;\checkmark\]

\[\text{By symmetry, the curve also passes through } (2,0), \text{ where } y' = \frac{1}{1-1/e} \approx 1.58198 \;\;\checkmark\]