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Given the equation $(x)^{y}=(y)^{x}$, we take the natural logarithm (ln) of both sides to simplify the exponents. $$ \ln(x^y) = \ln(y^x) $$ Using the power rule of logarithms, we get: $$ y \ln(x) = x \ln(y) $$
Now, we differentiate both sides of the equation $y \ln(x) = x \ln(y)$ with respect to $x$. We will use the product rule for differentiation, which states that $(uv)' = u'v + uv'$. $$ \frac{d}{dx} [y \ln(x)] = \frac{d}{dx} [x \ln(y)] $$ Applying the product rule: $$ \frac{dy}{dx} \ln(x) + y \cdot \frac{1}{x} = 1 \cdot \ln(y) + x \cdot \frac{1}{y} \frac{dy}{dx} $$ $$ \ln(x) \frac{dy}{dx} + \frac{y}{x} = \ln(y) + \frac{x}{y} \frac{dy}{dx} $$
We want to isolate $\frac{dy}{dx}$ on one side of the equation. Rearrange the terms: $$ \ln(x) \frac{dy}{dx} - \frac{x}{y} \frac{dy}{dx} = \ln(y) - \frac{y}{x} $$ Factor out $\frac{dy}{dx}$: $$ \frac{dy}{dx} \left( \ln(x) - \frac{x}{y} \right) = \ln(y) - \frac{y}{x} $$
Now, divide both sides by $\left( \ln(x) - \frac{x}{y} \right)$ to solve for $\frac{dy}{dx}$: $$ \frac{dy}{dx} = \frac{\ln(y) - \frac{y}{x}}{\ln(x) - \frac{x}{y}} $$ To simplify, multiply the numerator and denominator by $xy$: $$ \frac{dy}{dx} = \frac{xy \ln(y) - y^2}{xy \ln(x) - x^2} $$
Final Answer: $\frac{dy}{dx} = \frac{y(x\ln y - y)}{x(y\ln x - x)}$
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