Find frac{dy}{dx} for the function xy = e^{(x | vedclass

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(A) Given the function $xy = e^{(x-y)}$.Taking the natural logarithm on both sides:$\ln(xy) = \ln(e^{(x-y)})$$\ln x + \ln y = (x - y) \ln e$Since $\ln

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

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(A) Given the function $xy = e^{(x-y)}$.Taking the natural logarithm on both sides:$\ln(xy) = \ln(e^{(x-y)})$$\ln x + \ln y = (x - y) \ln e$Since $\ln e = 1$,we have:$\ln x + \ln y = x - y$Now,differentiate both sides with respect to $x$:$\frac{d}{dx}(\ln x) + \frac{d}{dx}(\ln y) = \frac{d}{dx}(x) - \frac{dy}{dx}$$\frac{1}{x} + \frac{1}{y} \frac{dy}{dx} = 1 - \frac{dy}{dx}$Rearrange the terms to group $\frac{dy}{dx}$:$\frac{1}{y} \frac{dy}{dx} + \frac{dy}{dx} = 1 - \frac{1}{x}$$\frac{dy}{dx} \left(\frac{1+y}{y}\right) = \frac{x-1}{x}$$\frac{dy}{dx} = \frac{y(x-1)}{x(y+1)}$

Find $\frac{dy}{dx}$ for the function $xy = e^{(x-y)}$.

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