If frac{x}{x-y} = log left(frac{a}{x-y}right) | vedclass

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(B) Given $\frac{x}{x-y} = \log a - \log(x-y)$.Rearranging,we get $\log(x-y) + \frac{x}{x-y} = \log a$.Differentiating both sides with respect to $x$:

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$\frac{2y - x}{y}$

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(B) Given $\frac{x}{x-y} = \log a - \log(x-y)$.Rearranging,we get $\log(x-y) + \frac{x}{x-y} = \log a$.Differentiating both sides with respect to $x$:$\frac{1}{x-y} \left(1 - \frac{dy}{dx}\right) + \frac{(x-y)(1) - x(1 - \frac{dy}{dx})}{(x-y)^2} = 0$.Multiply by $(x-y)^2$:$(x-y)(1 - \frac{dy}{dx}) + x - y - x + x \frac{dy}{dx} = 0$.$x - y - (x-y) \frac{dy}{dx} - y + x \frac{dy}{dx} = 0$.$x - 2y + \frac{dy}{dx} (x - (x-y)) = 0$.$x - 2y + y \frac{dy}{dx} = 0$.$y \frac{dy}{dx} = 2y - x$.Therefore,$\frac{dy}{dx} = \frac{2y - x}{y}$.

$If \frac{x}{x-y} = \log \left(\frac{a}{x-y}\right)$,then $\frac{dy}{dx} =$

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