If x^y=y^x, then x(x-y log x) frac{d y}{d x} | vedclass

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(A) Given equation is $x^y = y^x$. Taking natural logarithm on both sides, we get $y \log x = x \log y$. Differentiating both sides with respect to $x

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$y(y-x \log y)$

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(A) Given equation is $x^y = y^x$. Taking natural logarithm on both sides, we get $y \log x = x \log y$. Differentiating both sides with respect to $x$ using the product rule: $y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx} = x \cdot \frac{1}{y} \cdot \frac{dy}{dx} + \log y$. Rearranging the terms to isolate $\frac{dy}{dx}$: $\frac{dy}{dx} \left( \log x - \frac{x}{y} \right) = \log y - \frac{y}{x}$. $\frac{dy}{dx} \left( \frac{y \log x - x}{y} \right) = \frac{x \log y - y}{x}$. Multiplying both sides by $x$ and rearranging: $x \left( \frac{y \log x - x}{y} \right) \frac{dy}{dx} = x \log y - y$. Multiplying by $-1$ to match the required form: $x \left( \frac{x - y \log x}{y} \right) \frac{dy}{dx} = y - x \log y$. Therefore, $x(x - y \log x) \frac{dy}{dx} = y(y - x \log y)$.

If $x^y=y^x$, then $x(x-y \log x) \frac{d y}{d x}$ is equal to :

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