But there's two values of \(y\), no?

Calculus Level 4

If the positive value of \( \dfrac{dy}{dx} \) of the \(y = x^y\) at \(x = \sqrt2\) is equal to \( \dfrac{\sqrt A}{1 - \ln B} \), where \(A\) and \(B\) are positive integers, find \(A\times B\).

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