Question

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

• A.
• B.
• C. (2y-1)
• D.

Answer 1

The correct option is B

$\therefore$ y = $\sqrt{x+\sqrt{y+y}}$

$\Rightarrow$ $(y^2-x)=\sqrt{2y}$

or $(y^2-x{)}^2$ = 2y

Differentiating both sides w.r.t.x , then

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

$\therefore$ $\frac{dy}{dx}=\frac{(y^2-x)}{2y^3-2xy-1}$