For xy = e^{x-y},frac{dy}{dx} = . . . . . . | vedclass

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Question

(A) Given the equation $xy = e^{x-y}$.Taking the natural logarithm on both sides,we get:$\ln(xy) = \ln(e^{x-y})$$\ln x + \ln y = x - y$Differentiating

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the equation $xy = e^{x-y}$.Taking the natural logarithm on both sides,we get:$\ln(xy) = \ln(e^{x-y})$$\ln x + \ln y = x - y$Differentiating both sides with respect to $x$:$\frac{d}{dx}(\ln x) + \frac{d}{dx}(\ln y) = \frac{d}{dx}(x) - \frac{d}{dx}(y)$$\frac{1}{x} + \frac{1}{y} \frac{dy}{dx} = 1 - \frac{dy}{dx}$Rearranging the terms to collect $\frac{dy}{dx}$:$\frac{1}{y} \frac{dy}{dx} + \frac{dy}{dx} = 1 - \frac{1}{x}$$\frac{dy}{dx} (\frac{1}{y} + 1) = \frac{x-1}{x}$$\frac{dy}{dx} (\frac{1+y}{y}) = \frac{x-1}{x}$$\frac{dy}{dx} = \frac{y(x-1)}{x(y+1)}$Thus,the correct option is $A$.

For $xy = e^{x-y}$,$\frac{dy}{dx} =$ . . . . . .

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