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)

None of these

We have,
$y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+......\infty }}}}$
$\Rightarrow y^2=x+\sqrt{y+\sqrt{x+\sqrt{y+........\infty }}}$
$\Rightarrow y^2=x+\sqrt{y+y}$ ...... [From the definition of $y$]

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

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

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