Find frac{dy}{dx} for the function y^{x} = x^ | vedclass

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

Vedclass · Accepted Answer

$\frac{y}{x} \left( \frac{y - x \log y}{x - y \log x} \right)$

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

Find $\frac{dy}{dx}$ for the function $y^{x} = x^{y}$.

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