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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Answer

(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} = $

Similar Questions

Let $y = \sqrt{x + \sqrt{x + \sqrt{x + \dots \infty}}}$,then $\frac{dy}{dx} =$

If $x = e^{(y+e)^{(y+e)^{(y+\ldots \infty)}}}$,then $\frac{dy}{dx} = $

If $ y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}} $,then $ \frac{d y}{d x}= $

The derivative of $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots \infty}}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module