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

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

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} = $

Similar Questions

If $\sin x \sqrt{\cos y} - \cos y \sqrt{\sin x} = 0$,then $\frac{dy}{dx} = $

If $x^2 \tan ^{-1} \frac{y}{x}-y^2 \tan ^{-1} \frac{x}{y}=k$,then $\left(\frac{d y}{d x}\right)_{(1,1)}=$

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

Find $\frac{dy}{dx}$ for the equation $xy + y^2 = \tan x + y$.

If $\sin \left(\frac{x+y}{x-y}\right)=\tan \frac{\pi}{5}$,then $\frac{d y}{d x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module