If x 0 and x^y = e^{x-y},then frac{dy}{dx} | vedclass

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(B) Given the equation $x^y = e^{x-y}$.Taking the natural logarithm on both sides,we get:$\ln(x^y) = \ln(e^{x-y})$$y \ln x = x - y$Rearranging the ter

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

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(B) Given the equation $x^y = e^{x-y}$.Taking the natural logarithm on both sides,we get:$\ln(x^y) = \ln(e^{x-y})$$y \ln x = x - y$Rearranging the terms to isolate $y$:$y \ln x + y = x$$y(1 + \ln x) = x$$y = \frac{x}{1 + \ln x}$Now,differentiate with respect to $x$ using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:$\frac{dy}{dx} = \frac{(1)(1 + \ln x) - x(\frac{1}{x})}{(1 + \ln x)^2}$$\frac{dy}{dx} = \frac{1 + \ln x - 1}{(1 + \ln x)^2}$$\frac{dy}{dx} = \frac{\ln x}{(1 + \ln x)^2}$Thus,the correct option is $B$.

If $x > 0$ and $x^y = e^{x-y}$,then $\frac{dy}{dx}$ is equal to

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