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

Answer 1

Answer: (D)

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