If y = x^{x^{x...infty}}, then x (1 - y log x | vedclass

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(B) Given the equation $y = x^{x^{x...\infty}}$.Since the exponent is infinite,we can write $y = x^y$.Taking the natural logarithm on both sides,we ge

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$y^2$

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(B) Given the equation $y = x^{x^{x...\infty}}$.Since the exponent is infinite,we can write $y = x^y$.Taking the natural logarithm on both sides,we get $\log y = \log(x^y) = y \log x$.Differentiating both sides with respect to $x$,we get $\frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \log x + y \cdot \frac{1}{x}$.Multiplying the entire equation by $xy$,we get $x \frac{dy}{dx} = xy \log x \frac{dy}{dx} + y^2$.Rearranging the terms,we get $x \frac{dy}{dx} - xy \log x \frac{dy}{dx} = y^2$.Factoring out $x \frac{dy}{dx}$,we get $x \frac{dy}{dx} (1 - y \log x) = y^2$.Thus,$x (1 - y \log x) \frac{dy}{dx} = y^2$.

If $y = x^{x^{x...\infty}},$ then $x (1 - y \log x) \frac{dy}{dx} =$

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