If $xy = e^{x - y}$, then $\frac{dy}{dx}$ is equal to:
$\frac{\log x}{\log(x - y)}$
$\frac{e^x}{x^{x - y}}$
$\frac{\log x}{(1 + \log x)^2}$
$\frac{1}{y} - \frac{1}{x - y}$
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