If y = (sqrt{x})^{(sqrt{x})^{(sqrt{x})^{dotsi | vedclass

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(D) Given $y = (\sqrt{x})^{(\sqrt{x})^{(\sqrt{x})^{\dots\infty}}}$.Since the exponent is infinite,we can write $y = (\sqrt{x})^y$.Taking the natural l

Vedclass · Accepted Answer

None of these

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(D) Given $y = (\sqrt{x})^{(\sqrt{x})^{(\sqrt{x})^{\dots\infty}}}$.Since the exponent is infinite,we can write $y = (\sqrt{x})^y$.Taking the natural logarithm on both sides: $\log y = \log((\sqrt{x})^y) = y \log(\sqrt{x}) = y \log(x^{1/2}) = \frac{1}{2} y \log x$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left( \log x \cdot \frac{dy}{dx} + y \cdot \frac{1}{x} \right)$.Multiply by $2y$ to clear the fractions:$2 \frac{dy}{dx} = y \log x \frac{dy}{dx} + \frac{y^2}{x}$.Rearranging the terms to solve for $\frac{dy}{dx}$:$\frac{dy}{dx} (2 - y \log x) = \frac{y^2}{x}$.Therefore,$\frac{dy}{dx} = \frac{y^2}{x(2 - y \log x)}$.Comparing this result with the given options,none of the options $A, B,$ or $C$ match the derived expression. Thus,the correct option is $D$.

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

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