Question

If $y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+...\infty }}}$ show that $\frac{dy}{dx}=\frac{1}{x(2y-1)}$.

Answer 1

We have,

$y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x.....}}}$

$y=\sqrt{\log x+y}$ (squaring)

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

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

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

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

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

Hence proved.