If $y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \ldots \infty}}}$, then $\frac{d y}{d x}$ is equal to

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Aicharya · Accepted Answer

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} $

If $y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \ldots \infty}}}$, then $\frac{d y}{d x}$ is equal to

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