Question

If $y=\sqrt{\log x+2\sqrt{\log x+2\sqrt{\log x+\cdots \infty }}}$, then $\frac{dy}{dx}$ at $x=1$ is

• A. $0$
• B. $-1$
• C. $1$
• D. $-\frac{1}{2}$

Answer 1

The correct option is D $-\frac{1}{2}$

$y=\sqrt{\log x+2\sqrt{\log x+2\sqrt{\log x+\cdots \infty }}}$
$\Rightarrow y=\sqrt{\log x+2y}$
Squaring
$y^2=\log x+2y$
Diffrentiating
$2y{y}^{\prime }=\frac{1}{x}+2y^{\prime }$.......(i)
at $x=1,y=0$
From (i)
$0=\frac{1}{1}+2y^{\prime }\Rightarrow y^{\prime }{|}_{x=1}=\frac{1}{2}$