Question

Y=✓[sinx+✓{sinx+✓(sinx+.....inf.)}]​

Answer 1

Answer:

$y = \sqrt{ \sin(x) + \sqrt{ \sin(x) + \sqrt{ \sin(x) + ... \: to \: \infty } } } \\ y = \sqrt{ \sin(x) + y } \\ {y}^{2} = \sin(x) + y \\ On \: differentiating \: both \: sides, \: we \: get \\ 2y \frac{dy}{dx} = \cos(x) + \frac{dy}{dx} \\ (2y - 1) \frac{dy}{dx} = \cos(x) \\ \boxed{ \frac{dy}{dx} = \frac{ \cos(x) }{(2y - 1)} }✓$