Question

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

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

Answer 1

The correct option is C $\frac{1}{x(2y-1)}$

$y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\cdots \infty }}}$
$y=\sqrt{\log x+y}$
Squaring both the sides, we get
$y^2=\log x+y$
Differentiate w r to $x$
$2y\frac{dy}{dx}=\frac{1}{x}+\frac{dy}{dx}$
$2y\frac{dy}{dx}-\frac{dy}{dx}=\frac{1}{x}$
$\frac{dy}{dx}(2y-1)=\frac{1}{x}$
$\frac{dy}{dx}=\frac{1}{x(2y-1)}$