If e^{-y} cdot y = x,then frac{dy}{dx} is | vedclass

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(D) Given the equation $e^{-y} \cdot y = x$. Taking the derivative with respect to $x$ on both sides: $\frac{d}{dx}(y e^{-y}) = \frac{d}{dx}(x)$ Using

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$\frac{y}{x(1-y)}$

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(D) Given the equation $e^{-y} \cdot y = x$. Taking the derivative with respect to $x$ on both sides: $\frac{d}{dx}(y e^{-y}) = \frac{d}{dx}(x)$ Using the product rule: $y \cdot \frac{d}{dx}(e^{-y}) + e^{-y} \cdot \frac{dy}{dx} = 1$ $y \cdot (-e^{-y}) \frac{dy}{dx} + e^{-y} \frac{dy}{dx} = 1$ Factor out $\frac{dy}{dx}$: $\frac{dy}{dx} (e^{-y} - y e^{-y}) = 1$ $\frac{dy}{dx} e^{-y} (1 - y) = 1$ $\frac{dy}{dx} = \frac{1}{e^{-y}(1 - y)}$ Since $e^{-y} = \frac{x}{y}$,substitute this into the expression: $\frac{dy}{dx} = \frac{1}{(\frac{x}{y})(1 - y)} = \frac{y}{x(1 - y)}$.

If $e^{-y} \cdot y = x$,then $\frac{dy}{dx}$ is

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