If {x^y} = {y^x}, then frac{dy}{dx} = | vedclass

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(B) Given the equation ${x^y} = {y^x}$.Taking the natural logarithm on both sides,we get $y \log_e x = x \log_e y$.Differentiating both sides with res

Vedclass · Accepted Answer

$\frac{y(y - x\log_e y)}{x(x - y\log_e x)}$

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(B) Given the equation ${x^y} = {y^x}$.Taking the natural logarithm on both sides,we get $y \log_e x = x \log_e y$.Differentiating both sides with respect to $x$ using the product rule:$\frac{d}{dx}(y \log_e x) = \frac{d}{dx}(x \log_e y)$$\frac{dy}{dx} \log_e x + y \cdot \frac{1}{x} = 1 \cdot \log_e y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx}$Rearranging the terms to isolate $\frac{dy}{dx}$:$\frac{dy}{dx} \log_e x - \frac{x}{y} \frac{dy}{dx} = \log_e y - \frac{y}{x}$$\frac{dy}{dx} \left( \log_e x - \frac{x}{y} \right) = \log_e y - \frac{y}{x}$$\frac{dy}{dx} \left( \frac{y \log_e x - x}{y} \right) = \frac{x \log_e y - y}{x}$$\frac{dy}{dx} = \frac{y(x \log_e y - y)}{x(y \log_e x - x)}$Multiplying numerator and denominator by $-1$,we get $\frac{dy}{dx} = \frac{y(y - x \log_e y)}{x(x - y \log_e x)}$.

If ${x^y} = {y^x},$ then $\frac{dy}{dx} = $

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