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=(x/(a+(x/(b+z)))) ⇒ z=((−b+(√((4x+ab)b)))/2) y=x+z ⇒ y=x+((−b+(√((4x+ab)b)))/2) ((d[x+((−b+(√((4x+ab)b)))/2)])/dx)=1+((√b)/( (√((4x+ab)a))))](https://www.tinkutara.com/question/Q192902.png)
$$\mathrm{Assuming}\:{a},\:{b}\:>\mathrm{0} \\ $$$${z}=\frac{{x}}{{a}+\frac{{x}}{{b}+{z}}}\:\Rightarrow\:{z}=\frac{−{b}+\sqrt{\left(\mathrm{4}{x}+{ab}\right){b}}}{\mathrm{2}} \\ $$$${y}={x}+{z}\:\Rightarrow\:{y}={x}+\frac{−{b}+\sqrt{\left(\mathrm{4}{x}+{ab}\right){b}}}{\mathrm{2}} \\ $$$$\frac{{d}\left[{x}+\frac{−{b}+\sqrt{\left(\mathrm{4}{x}+{ab}\right){b}}}{\mathrm{2}}\right]}{{dx}}=\mathrm{1}+\frac{\sqrt{{b}}}{\:\sqrt{\left(\mathrm{4}{x}+{ab}\right){a}}} \\ $$
Answered by horsebrand11 last updated on 01/Jun/23
$$\:{y}−{x}\:=\:\frac{{x}}{{a}+\frac{{x}}{{b}+{y}−{x}}} \\ $$$$\:{y}−{x}=\frac{{bx}+{xy}−{x}^{\mathrm{2}} }{{ab}+{ay}−{ax}+{x}} \\ $$$$\:\left({y}−{x}\right)\left({ab}+{ay}−{ax}+{x}\right)={bx}+{xy}−{x}^{\mathrm{2}} \\ $$$$\:{aby}+{ay}^{\mathrm{2}} −{axy}+\cancel{{xy}}−{abx}−{axy}+{ax}^{\mathrm{2}} −\cancel{{x}^{\mathrm{2}} } \\ $$$$\:={bx}+\cancel{{xy}}−\cancel{{x}^{\mathrm{2}} } \\ $$$$\:{ay}^{\mathrm{2}} −\mathrm{2}{axy}+{aby}−{abx}−{bx}+{ax}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\mathrm{2}{ayy}'−\mathrm{2}{ay}−\mathrm{2}{axy}'−{ab}−{b}+\mathrm{2}{ax}=\mathrm{0} \\ $$$$\:{y}'\left(\mathrm{2}{ay}−\mathrm{2}{ax}\right)=\mathrm{2}{ay}−\mathrm{2}{ax}+{ab}+{b} \\ $$$$\:\Rightarrow\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}{ay}−\mathrm{2}{ax}+{ab}+{b}}{\mathrm{2}{ay}−\mathrm{2}{ax}}\: \\ $$