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Q. If $y = \sqrt{x+\sqrt{x+\sqrt{x+.......\infty}}}$, then $\frac{dy}{dx}$ is equal to

Limits and Derivatives

Solution:

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