Q69 / 120 · VITEEE 2010 · Mathematics
If $y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+\ldots \infty}}}}$, then $\frac{d y}{d x}$ is equal to
- A$\frac{y+x}{y^{2}-2 x}$
- B$\frac{y^{3}-x}{2 y^{2}-2 x y-1}$
- C$\frac{y^{3}+x}{2 y^{2}-x}$
- Correct answerDNone of these
Solution
$\begin{aligned} y &=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+\ldots \ldots \infty}}}} \\ \Rightarrow y^{2} &=x+\sqrt{y+\sqrt{x+\sqrt{y+\ldots \ldots \infty}}} \\ \Rightarrow y^{2} &=x+\sqrt{y+y} \Rightarrow y^{2}=x+\sqrt{2 y} \\ \Rightarrow &\left(y^{2}-x\right)^{2}=2 y \end{aligned}$
On differentiating both sides w.r.t. $x$, we get
$$
\begin{array}{l}
2\left(y^{2}-x\right)\left(2 y \frac{d y}{d x}-1\right)=2 \frac{d y}{d x} \\
\Rightarrow 2\left(y^{3}-x y\right) \frac{d y}{d x}-\left(y^{2}-x\right)=\frac{d y}{d x} \\
\Rightarrow\left(2 y^{3}-2 x y-1\right) \frac{d y}{d x}=y^{2}-x \\
\Rightarrow \frac{d y}{d x}=\frac{y^{2}-x}{2 y^{3}-2 x y-1}
\end{array}
$$
Hence, the correct answer is option: 'D'.