Question

differenciate below
$\sqrt{x + \sqrt{x + \sqrt{x...} } }$
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Answer 1

Answer:

Your answer is 1/(2y-1)

Step-by-step explanation:

$let \: y = \sqrt{x + \sqrt{x} } \\ then \: y = \sqrt{x + y} \\ {y}^{2} = (x + y) \\ \: now \: differentiate \: with \: respect \: to \: x \\ \frac{dy}{dx} \times {y}^{2} = \frac{d}{dx} (x + y) \\ = > 2y \times \frac{dy}{dx} = 1 + \frac{dy}{dx} \\ = > (2y - 1) \frac{dy}{dx} = 1 \\ = > \frac{dy}{dx} = \frac{1}{(2y - 1)}$

hope it will help you ☺️