If log (x+y)=log (xy)+3,then frac{dy}{dx}= | vedclass

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(C) Given the equation: $\log (x+y) = \log (xy) + 3$ Using the property of logarithms $\log a - \log b = \log (\frac{a}{b})$,we get: $\log (x+y) - \lo

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$-\left(\frac{y}{x}\right)^2$

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(C) Given the equation: $\log (x+y) = \log (xy) + 3$ Using the property of logarithms $\log a - \log b = \log (\frac{a}{b})$,we get: $\log (x+y) - \log (xy) = 3$ $\log \left(\frac{x+y}{xy}\right) = 3$ Converting from logarithmic to exponential form: $\frac{x+y}{xy} = e^3$ $\frac{x}{xy} + \frac{y}{xy} = e^3$ $\frac{1}{y} + \frac{1}{x} = e^3$ Differentiating both sides with respect to $x$: $\frac{d}{dx} (y^{-1}) + \frac{d}{dx} (x^{-1}) = \frac{d}{dx} (e^3)$ $-y^{-2} \frac{dy}{dx} - x^{-2} = 0$ $-\frac{1}{y^2} \frac{dy}{dx} = \frac{1}{x^2}$ $\frac{dy}{dx} = -\frac{y^2}{x^2} = -\left(\frac{y}{x}\right)^2$

If $\log (x+y)=\log (xy)+3$,then $\frac{dy}{dx}=$

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