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

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(A) Given $\log y = y^{\log x}$.Taking $\log$ on both sides:$\log(\log y) = \log(y^{\log x}) = \log x \cdot \log y$.Differentiating both sides with re

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$\frac{y(\log y)^2}{x(1-\log x \log y)}$

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(A) Given $\log y = y^{\log x}$.Taking $\log$ on both sides:$\log(\log y) = \log(y^{\log x}) = \log x \cdot \log y$.Differentiating both sides with respect to $x$:$\frac{1}{\log y} \cdot \frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}(\log x) \cdot \log y + \log x \cdot \frac{d}{dx}(\log y)$.$\frac{1}{y \log y} \cdot \frac{dy}{dx} = \frac{\log y}{x} + \frac{\log x}{y} \cdot \frac{dy}{dx}$.Rearranging the terms to isolate $\frac{dy}{dx}$:$\frac{dy}{dx} \left( \frac{1}{y \log y} - \frac{\log x}{y} \right) = \frac{\log y}{x}$.$\frac{dy}{dx} \left( \frac{1 - \log x \log y}{y \log y} \right) = \frac{\log y}{x}$.$\frac{dy}{dx} = \frac{\log y}{x} \cdot \frac{y \log y}{1 - \log x \log y}$.$\frac{dy}{dx} = \frac{y(\log y)^2}{x(1 - \log x \log y)}$.

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

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