If x = y^{x^{y^{x^{y^{x = dots infty}}}}},the | vedclass

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(A) Given the infinite expression $x = y^{x^{y^{x^{\dots}}}}$,we can write it as $x = y^{x^x}$.Taking the natural logarithm on both sides,we get $\ln

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(A) Given the infinite expression $x = y^{x^{y^{x^{\dots}}}}$,we can write it as $x = y^{x^x}$.Taking the natural logarithm on both sides,we get $\ln x = x^x \ln y$.At $x = 1$,we have $\ln 1 = 1^1 \ln y$,which implies $0 = \ln y$,so $y = 1$.Differentiating both sides with respect to $x$ using the product rule and chain rule:$\frac{d}{dx}(\ln x) = \frac{d}{dx}(x^x \ln y)$$\frac{1}{x} = \frac{d}{dx}(x^x) \cdot \ln y + x^x \cdot \frac{d}{dx}(\ln y)$Since $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$,we have:$\frac{1}{x} = x^x(1 + \ln x) \ln y + x^x \cdot \frac{1}{y} y'$Substituting $x = 1$ and $y = 1$:$\frac{1}{1} = 1^1(1 + \ln 1) \ln 1 + 1^1 \cdot \frac{1}{1} y'$$1 = 1(1 + 0)(0) + 1 \cdot y'$$1 = 0 + y'$Therefore,$y' = 1$.

If $x = y^{x^{y^{x^{y^{x = \dots \infty}}}}}$,then $y'$ at $x=1$ is

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