Question

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

• A. $\frac{y^2+x}{2y^3-2xy-1}$
• B. $\frac{y^2-x}{2y^3-2xy-1}$
• C. $\frac{y^2-x}{2y^4-2xy+1}$
• D. $\frac{y^2}{y^3-2xy+1}$

Answer 1

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

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

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

$\therefore y=\sqrt{x+\sqrt{2y}}$

$\therefore y^2=x+\sqrt{2y}$

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

$\therefore y^4+x^2-2x{y}^2=4y^2$

Now differentiating with respect to x

$\therefore 4y^3\frac{dy}{dx}+2x-2y^2-4xy\frac{dy}{dx}=8y\frac{dy}{dx}$

$\therefore \frac{dy}{dx}(4y^3-4xy-8y)=2y^2-2x$

$\therefore \frac{dy}{dx}=\frac{2y^2-2x}{4y^3-4xy-8y}=\frac{y^2-x}{2y^3-2xy-4y}$

Hence $\frac{dy}{dx}=\frac{y^2-x}{2y^3-2xy-4y}$