Question

If $y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+}}}}...\infty$ then $\frac{dy}{dx}$ is equal to

• A. $\frac{x}{2y-1}$
• B. $\frac{x}{2y+1}$
• C. $\frac{1}{x(2y-1)}$
• D. $\frac{1}{x(1-2y)}$

Answer 1

Answer: (C)

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

Given, $y=\sqrt{\log x+y}$ ....... $[\because$ $y=\sqrt{\log x+\sqrt{\log x+}}...]$

$\Rightarrow y^2=\log x+y$

On differentiating w.r.t.x. we get

$2y\frac{dy}{dx}=\frac{1}{x}+\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{x(2y-1)}$