Differentiate: $x y = e^{(x - y)}$

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Solution

$x y = e^{(x - y)}$
$ln x y = (x - y) ln e$
$ln x y = x - y$
$\Rightarrow \frac{1}{x y} [x \frac{d y}{d x} + y] = 1 - \frac{d y}{d x}$
$\Rightarrow (\frac{1}{y} + 1) \frac{d y}{d x} = 1 - \frac{1}{x}$
$\Rightarrow \frac{d y}{d x} = \frac{\frac{x - 1}{x}}{\frac{y + 1}{y}} = \frac{y (x - 1)}{x (y + 1)}$

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