Question

If $y=\sqrt{x+\sqrt{x+\sqrt{x}}}+..\infty$ then $\frac{dy}{dx}=$

Answer 1

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

here, $\sqrt{x+\sqrt{x+\sqrt{x+......\infty }}}=\sqrt{x+y}$

so, $y=\sqrt{x+y}$

take the square on both the side we get,

$y^2=x+y$...........(2)

now, differentiate equation 2 w.r.t. $x$

we get

$2y\frac{dy}{dx}=1+\frac{dy}{dx}$

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

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

so ,

$\frac{dy}{dx}=\frac{1}{2\sqrt{x+\sqrt{x}+\sqrt{x}+.....\infty }}-1$