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

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(C) Given $x^y = e^{x-y}$.Taking logarithm on both sides,we get:$y \log x = x - y$$y(1 + \log x) = x$$y = \frac{x}{1 + \log x} \quad \dots(1)$Differen

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$\frac{\log x}{(1+\log x)^2}$

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(C) Given $x^y = e^{x-y}$.Taking logarithm on both sides,we get:$y \log x = x - y$$y(1 + \log x) = x$$y = \frac{x}{1 + \log x} \quad \dots(1)$Differentiating both sides with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{(1 + \log x) \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(1 + \log x)}{(1 + \log x)^2}$$\frac{dy}{dx} = \frac{(1 + \log x)(1) - x \cdot (\frac{1}{x})}{(1 + \log x)^2}$$\frac{dy}{dx} = \frac{1 + \log x - 1}{(1 + \log x)^2}$$\frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2}$

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

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