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 te

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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$$y = \frac{x}{1 + \ln x}$Now, differentiate $y$ with respect to $x$ using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$:Let $u = x$ and $v = 1 + \ln x$. Then $u' = 1$ and $v' = \frac{1}{x}$.$\frac{dy}{dx} = \frac{(1 + \ln x)(1) - x(\frac{1}{x})}{(1 + \ln x)^2}$$\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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