Question

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

Answer 1

Given,

$y=\sqrt{logx+\sqrt{logx+\sqrt{logx}}}+$_____ $\infty$.......(1)

Squaring both sides we get,

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

Differentiating both sides we get,

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

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