Question

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

Answer 1

Given,

$y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+......\infty }}}$.......(1).

Squaring both sides we've,

$y^2=\log x+$ $\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+......\infty }}}$

or, $y^2=\log x+y$ [ Using (1)]

Now differentiating both sides with respect to $x$ we get,

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

or, $\frac{dy}{dx}=\frac{1}{x(2y-1)}$.