If {x^y} cdot {y^x} = 1,then {{dy} over {dx}} | vedclass

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(B) Given the equation: ${x^y} \cdot {y^x} = 1$. Taking the natural logarithm on both sides: $\log({x^y} \cdot {y^x}) = \log(1)$. Using logarithmic pr

Vedclass · Accepted Answer

$ - {{y(y + x\log y)} \over {x(x + y\log x)}}$

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(B) Given the equation: ${x^y} \cdot {y^x} = 1$. Taking the natural logarithm on both sides: $\log({x^y} \cdot {y^x}) = \log(1)$. Using logarithmic properties: $y \log x + x \log y = 0$. Now,differentiate both sides with respect to $x$: $\frac{d}{dx}(y \log x) + \frac{d}{dx}(x \log y) = 0$. Applying the product rule: $(y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx}) + (x \cdot \frac{1}{y} \cdot \frac{dy}{dx} + \log y \cdot 1) = 0$. Rearranging the terms to isolate $\frac{dy}{dx}$: $\frac{dy}{dx} (\log x + \frac{x}{y}) = - (\frac{y}{x} + \log y)$. $\frac{dy}{dx} (\frac{y \log x + x}{y}) = - (\frac{y + x \log y}{x})$. Therefore,$\frac{dy}{dx} = - \frac{y(y + x \log y)}{x(x + y \log x)}$.

If ${x^y} \cdot {y^x} = 1$,then ${{dy} \over {dx}} = $

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