Question

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

• A. $\frac{y^2-x}{2y^3-2xy-1}$
• B. $\frac{x^2-x}{2x^3-2xy-1}$
• C. $\frac{x^2-x}{2x^3-2x{y}^2-1}$
• D. None of these

Answer 1

The correct option is A $\frac{y^2-x}{2y^3-2xy-1}$

$y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+\dots \infty }}}}$
$=\sqrt{x+\sqrt{y+y}}$
or $y^2=x+\sqrt{2y}$
Differentiating w.r.t. $x$, we get
or $2y\frac{dy}{dx}=1+\frac{1}{\sqrt{2y}}\times \frac{dy}{dx}$
or $\frac{dy}{dx}[2y-\frac{1}{\sqrt{2y}}]=1$
or $\frac{dy}{dx}=\frac{\sqrt{2y}}{2y\sqrt{2y}-1}$
$=\frac{y^2-x}{2y^3-2xy-1}$