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y = √(sinx + √(sinx + √(sinx + ... ∞ ))) Square on both sides:
y² = sinx + √(sinx + √(sinx + ... ∞ )))
y² = sinx + y
y² - y = sinx
Differentiate both sides w.r.t x:
=> d(y² - y)/dx = d(sinx)/dx
=> d(y² - y)/dy × dy/dx = cosx
=> (2y - 1) × dy/dz = cosx
=> dy/dx = cosx/(2y - 1) proved
Line2: √(sinx + √(sinx + ... ∞ ))) is replaced with y as both are equal.
Line8: using chain rule
d(y² - y)/dx = d(y² - y)/dy * dy/dx
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