# 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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