If y = x^{x^{x^{dotsinfty}}},then frac{dy}{dx | vedclass

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(B) Given the infinite power tower $y = x^{x^{x^{\dots\infty}}}$.Since the exponent repeats infinitely,we can write $y = x^y$.Taking the natural logar

Vedclass · Accepted Answer

$\frac{y^2}{x(1 - y \log x)}$

Vedclass · Answer

(B) Given the infinite power tower $y = x^{x^{x^{\dots\infty}}}$.Since the exponent repeats infinitely,we can write $y = x^y$.Taking the natural logarithm on both sides,we get $\log y = y \log x$.Differentiating both sides with respect to $x$ using the product rule on the right side:$\frac{1}{y} \frac{dy}{dx} = y \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{dy}{dx}$.$\frac{1}{y} \frac{dy}{dx} = \frac{y}{x} + \log x \frac{dy}{dx}$.Rearranging the terms to isolate $\frac{dy}{dx}$:$\frac{dy}{dx} (\frac{1}{y} - \log x) = \frac{y}{x}$.$\frac{dy}{dx} (\frac{1 - y \log x}{y}) = \frac{y}{x}$.Therefore,$\frac{dy}{dx} = \frac{y^2}{x(1 - y \log x)}$.

If $y = x^{x^{x^{\dots\infty}}}$,then $\frac{dy}{dx} = $

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