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

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(D) Given,$x^{x}=y^{y}$.Taking the natural logarithm on both sides,we get:$x \log x = y \log y$.Differentiating both sides with respect to $x$ using t

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$\frac{1+\log x}{1+\log y}$

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(D) Given,$x^{x}=y^{y}$.Taking the natural logarithm on both sides,we get:$x \log x = y \log y$.Differentiating both sides with respect to $x$ using the product rule:$\frac{d}{dx}(x \log x) = \frac{d}{dx}(y \log y)$.Applying the product rule $(uv)' = u'v + uv'$:$(1 \cdot \log x + x \cdot \frac{1}{x}) = (1 \cdot \log y + y \cdot \frac{1}{y} \cdot \frac{dy}{dx})$.$(\log x + 1) = (\log y + 1) \frac{dy}{dx}$.Therefore,$\frac{dy}{dx} = \frac{1+\log x}{1+\log y}$.

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

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