If (x e)^{y}=e^{x},then frac{d y}{d x} is | vedclass

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(A) Given equation is $(x e)^{y}=e^{x}$.Taking natural logarithm on both sides,we get:$y \log(x e) = x \log e$Since $\log(x e) = \log x + \log e$ and

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

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(A) Given equation is $(x e)^{y}=e^{x}$.Taking natural logarithm on both sides,we get:$y \log(x e) = x \log e$Since $\log(x e) = \log x + \log e$ and $\log e = 1$,we have:$y(\log x + 1) = x$$y = \frac{x}{\log x + 1}$Differentiating with respect to $x$ using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$:$\frac{dy}{dx} = \frac{(\log x + 1)(1) - x(\frac{1}{x} + 0)}{(\log x + 1)^2}$$\frac{dy}{dx} = \frac{\log x + 1 - 1}{(\log x + 1)^2}$$\frac{dy}{dx} = \frac{\log x}{(\log x + 1)^2}$

If $(x e)^{y}=e^{x}$,then $\frac{d y}{d x}$ is

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