If y = log_y x,then frac{dy}{dx} is equal to | vedclass

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(C) Given the equation $y = \log_y x$. Using the change of base formula,we can write this as $y = \frac{\log_e x}{\log_e y}$. Rearranging the terms,we

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$\frac{1}{x(1 + \log y)}$

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(C) Given the equation $y = \log_y x$. Using the change of base formula,we can write this as $y = \frac{\log_e x}{\log_e y}$. Rearranging the terms,we get $y \cdot \log_e y = \log_e x$. Differentiating both sides with respect to $x$ using the product rule on the left side: $\frac{d}{dx}(y \cdot \log_e y) = \frac{d}{dx}(\log_e x)$ $y \cdot \frac{d}{dx}(\log_e y) + \log_e y \cdot \frac{dy}{dx} = \frac{1}{x}$ $y \cdot (\frac{1}{y} \cdot \frac{dy}{dx}) + \log_e y \cdot \frac{dy}{dx} = \frac{1}{x}$ $1 \cdot \frac{dy}{dx} + \log_e y \cdot \frac{dy}{dx} = \frac{1}{x}$ $(1 + \log_e y) \frac{dy}{dx} = \frac{1}{x}$ Therefore,$\frac{dy}{dx} = \frac{1}{x(1 + \log_e y)}$.

If $y = \log_y x$,then $\frac{dy}{dx}$ is equal to

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