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If $y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \ldots \infty}}}$, then $\frac{d y}{d x}$ is equal to
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1280 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2 y-1}$
We have,
$$
\begin{aligned}
y & =\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}} \\
y^2 & =x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} \\
y^2 & =x+y \\
y^2-y & =x
\end{aligned}
$$
On differentiating w.r.t. $x$, we get
$$
\begin{aligned}
(2 y-1) \frac{d y}{d x} & =1 \\
\frac{d y}{d x} & =\frac{1}{2 y-1}
\end{aligned}
$$
$$
\begin{aligned}
y & =\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}} \\
y^2 & =x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} \\
y^2 & =x+y \\
y^2-y & =x
\end{aligned}
$$
On differentiating w.r.t. $x$, we get
$$
\begin{aligned}
(2 y-1) \frac{d y}{d x} & =1 \\
\frac{d y}{d x} & =\frac{1}{2 y-1}
\end{aligned}
$$
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