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Question Number 192901 by cortano12 last updated on 30/May/23

	Answered by Frix last updated on 30/May/23

	$Assuming a, b > 0$ $z = \frac{x}{a + \frac{x}{b + z}} \Rightarrow z = \frac{- b + \sqrt{(4 x + a b) b}}{2}$ $y = x + z \Rightarrow y = x + \frac{- b + \sqrt{(4 x + a b) b}}{2}$ $\frac{d [x + \frac{- b + \sqrt{(4 x + a b) b}}{2}]}{d x} = 1 + \frac{\sqrt{b}}{\sqrt{(4 x + a b) a}}$
	Answered by horsebrand11 last updated on 01/Jun/23

	$y - x = \frac{x}{a + \frac{x}{b + y - x}}$ $y - x = \frac{b x + x y - x^{2}}{a b + a y - a x + x}$ $(y - x) (a b + a y - a x + x) = b x + x y - x^{2}$ $a b y + {a y}^{2} - a x y + x y - a b x - a x y + {a x}^{2} - x^{2}$ $= b x + x y - x^{2}$ ${a y}^{2} - 2 a x y + a b y - a b x - b x + {a x}^{2} = 0$ $2 {a y y}^{'} - 2 a y - 2 {a x y}^{'} - a b - b + 2 a x = 0$ $y^{'} (2 a y - 2 a x) = 2 a y - 2 a x + a b + b$ $\Rightarrow \frac{d y}{d x} = \frac{2 a y - 2 a x + a b + b}{2 a y - 2 a x}$
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