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Given the equation $xe^y = 1$, we need to find $\frac{dy}{dx}$. We will use implicit differentiation with respect to $x$.
Differentiating both sides of $xe^y = 1$ with respect to $x$, we get: $$\frac{d}{dx}(xe^y) = \frac{d}{dx}(1)$$ Using the product rule on the left side: $$\frac{d}{dx}(x) \cdot e^y + x \cdot \frac{d}{dx}(e^y) = 0$$ $$1 \cdot e^y + x \cdot e^y \cdot \frac{dy}{dx} = 0$$
Now, we solve for $\frac{dy}{dx}$: $$e^y + xe^y \frac{dy}{dx} = 0$$ $$xe^y \frac{dy}{dx} = -e^y$$ $$\frac{dy}{dx} = -\frac{e^y}{xe^y}$$ $$\frac{dy}{dx} = -\frac{1}{x}$$
We need to find the value of $\frac{dy}{dx}$ at $x=1$. $$\frac{dy}{dx}\Big|_{x=1} = -\frac{1}{1} = -1$$
Final Answer: -1
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