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If $y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+\ldots \infty}}}}$, then $\frac{d y}{d x}=$
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Verified Answer
The correct answer is:
$\frac{y^2-x}{2 y^3-2 x y-1}$
We have
$$
\begin{aligned}
& y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+\ldots \infty}}}} \\
& y=\sqrt{x+\sqrt{y+y}} \Rightarrow y^2=x+\sqrt{2 y} \\
\Rightarrow & y^2-x=\sqrt{2 y} \Rightarrow\left(y^2-x\right)^2=2 y \\
\Rightarrow & y^4-2 x y^2+x^2=2 y \\
\Rightarrow & y^4-2 x y^2-2 y+x^2=0
\end{aligned}
$$
On differentiating w.r.t. to $x$, we get
$$
\begin{aligned}
& 4 y^3 \frac{d y}{d x}-2 y^2-4 y x \frac{d y}{d x}-\frac{2 d y}{d x}+2 x=0 \\
& \frac{d y}{d x}\left(4 y^3-4 x y-2\right)=-2 x+2 y^2 \\
\Rightarrow \quad & \frac{d y}{d x}=\frac{y^2-x}{2 y^3-2 x y-1}
\end{aligned}
$$
$$
\begin{aligned}
& y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+\ldots \infty}}}} \\
& y=\sqrt{x+\sqrt{y+y}} \Rightarrow y^2=x+\sqrt{2 y} \\
\Rightarrow & y^2-x=\sqrt{2 y} \Rightarrow\left(y^2-x\right)^2=2 y \\
\Rightarrow & y^4-2 x y^2+x^2=2 y \\
\Rightarrow & y^4-2 x y^2-2 y+x^2=0
\end{aligned}
$$
On differentiating w.r.t. to $x$, we get
$$
\begin{aligned}
& 4 y^3 \frac{d y}{d x}-2 y^2-4 y x \frac{d y}{d x}-\frac{2 d y}{d x}+2 x=0 \\
& \frac{d y}{d x}\left(4 y^3-4 x y-2\right)=-2 x+2 y^2 \\
\Rightarrow \quad & \frac{d y}{d x}=\frac{y^2-x}{2 y^3-2 x y-1}
\end{aligned}
$$
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