Given the equation $e^{-x} + e^{-y} = 2$. Differentiating both sides with respect to $x$, we apply the chain rule:
$$\frac{d}{dx}(e^{-x}) + \frac{d}{dx}(e^{-y}) = \frac{d}{dx}(2)$$ $$-e^{-x} - e^{-y} \cdot \frac{dy}{dx} = 0$$Rearrange the equation to solve for $\frac{dy}{dx}$:
$$-e^{-y} \cdot \frac{dy}{dx} = e^{-x}$$ $$\frac{dy}{dx} = -\frac{e^{-x}}{e^{-y}}$$Using the laws of exponents, specifically $\frac{e^a}{e^b} = e^{a-b}$:
$$\frac{dy}{dx} = -e^{-x - (-y)}$$ $$\frac{dy}{dx} = -e^{y-x}$$Final Answer: $-e^{y-x}$
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