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If xy=ex−yx^y = e^{x-y}, then dydx=\frac{dy}{dx} =

Solution

✅ Correct Option: 2

Taking log: ylog⁡x=x−yy\log x = x - y, so y=x1+log⁡xy = \frac{x}{1+\log x}. Differentiating: dydx=(1+log⁡x)−x⋅1x(1+log⁡x)2=log⁡x(1+log⁡x)2\frac{dy}{dx} = \frac{(1+\log x) - x \cdot \frac{1}{x}}{(1+\log x)^2} = \frac{\log x}{(1+\log x)^2}.

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