Question

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

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

Answer 1

Answer: (D)

$y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+...\infty }}}}$
$\Rightarrow y^2=x+\sqrt{y+\sqrt{x+\sqrt{y+...\infty }}}$
$\Rightarrow y^2=x+\sqrt{y+y}$
$\Rightarrow y^2=x+\sqrt{2y}$
$\Rightarrow {\left(y^2-x\right)}^2=2y$
On differentiating both sides w.r.t. $x$, we get
$2\left(y^2-x\right)\left(2y\frac{dy}{dx}-1\right)=2\frac{dy}{dx}$
$\Rightarrow 2\left(y^3-xy\right)\frac{dy}{dx}-\left(y^2-x\right)=\frac{dy}{dx}$
$\Rightarrow \left(2y^3-2xy-1\right)\frac{dy}{dx}=y^2-x$
$\Rightarrow \frac{dy}{dx}=\frac{y^2-x}{2y^3-2xy-1}$