If y = sqrt{log x + sqrt{log x + sqrt{log x + | vedclass

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Question

(C) Given the equation $y = \sqrt{\log x + \sqrt{\log x + \sqrt{\log x + \dots \infty}}}$.Since the series is infinite,we can write $y = \sqrt{\log x

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given the equation $y = \sqrt{\log x + \sqrt{\log x + \sqrt{\log x + \dots \infty}}}$.Since the series is infinite,we can write $y = \sqrt{\log x + y}$.Squaring both sides,we get $y^2 = \log x + y$.Differentiating both sides with respect to $x$:$\frac{d}{dx}(y^2) = \frac{d}{dx}(\log x + y)$$2y \frac{dy}{dx} = \frac{1}{x} + \frac{dy}{dx}$.Rearranging the terms to solve for $\frac{dy}{dx}$:$2y \frac{dy}{dx} - \frac{dy}{dx} = \frac{1}{x}$$\frac{dy}{dx}(2y - 1) = \frac{1}{x}$$\frac{dy}{dx} = \frac{1}{x(2y - 1)}$.Thus,the correct option is $C$.

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

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