Find absolute value of $d y / d x$ at $x = 1$ if $y^{x} . x^{y} = 1$

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Solution

$x^{y} . y^{x} = 1$
Taking logarithm, we get, $x log y + y log x = log 1 = 0.$
Differentiating w.r.t. $x$ ,
we get
$x \cdot \frac{1}{y} \frac{d y}{d x} + 1. log y + y . \frac{1}{x} + \frac{d y}{d x} log x = 0$
$∴ \frac{d y}{d x} = - \frac{log y + y / x}{x / y + log x} ∴ | y^{'} (1) | = | - 1 | = 1$

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