If y=cosx+cosx+⋯+∞, then dydx=

If $y = \sqrt{\cos x + \sqrt{\cos x + \dots + \infty}}$ , then $\frac{d y}{d x} =$

A
$\frac{\sin x}{2 y - 1}$
B
$\frac{\cos x}{2 y - 1}$
C
$\frac{\cos x}{1 - 2 y}$
D
$\frac{\sin x}{1 - 2 y}$

Solution:

$y = \sqrt{\cos x + y}$

Squaring on both sides

$y^{2} = \cos x + y$

$2 y . \frac{d y}{d x} = - \sin x + \frac{d y}{d x}$

$(2 y - 1) \frac{d y}{d x} = - \sin x$

$\frac{d y}{d x} = \frac{- \sin x}{2 y - 1}$

$\frac{d y}{d x} = \frac{\sin x}{1 - 2 y}$

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