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Question-145163




Question Number 145163 by mim24 last updated on 02/Jul/21
Answered by mathmax by abdo last updated on 02/Jul/21
y=(√(x+(√(x+(√(x+...+∞)))))) ⇒y=(√(x+y)) ⇒y^2  =x+y ⇒  y^2 −y−x=0 →Δ=1+4x ⇒y=((1+(√(1+4x)))/2) or y=((1−(√(1+4x)))/2)  but y≥0 ⇒y=((1+(√(4x+1)))/2) ⇒(dy/dx)=(1/2)×(4/(2(√(4x+1))))=(1/( (√(4x+1)))) ⇒  (dy/dx)=(1/( (√(4x+1))))
$$\mathrm{y}=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+…+\infty}}}\:\Rightarrow\mathrm{y}=\sqrt{\mathrm{x}+\mathrm{y}}\:\Rightarrow\mathrm{y}^{\mathrm{2}} \:=\mathrm{x}+\mathrm{y}\:\Rightarrow \\ $$$$\mathrm{y}^{\mathrm{2}} −\mathrm{y}−\mathrm{x}=\mathrm{0}\:\rightarrow\Delta=\mathrm{1}+\mathrm{4x}\:\Rightarrow\mathrm{y}=\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4x}}}{\mathrm{2}}\:\mathrm{or}\:\mathrm{y}=\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4x}}}{\mathrm{2}} \\ $$$$\mathrm{but}\:\mathrm{y}\geqslant\mathrm{0}\:\Rightarrow\mathrm{y}=\frac{\mathrm{1}+\sqrt{\mathrm{4x}+\mathrm{1}}}{\mathrm{2}}\:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{4x}+\mathrm{1}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{4x}+\mathrm{1}}}\:\Rightarrow \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{4x}+\mathrm{1}}} \\ $$
Answered by puissant last updated on 02/Jul/21
Cherchons (dy/dx)  y=(√(x+(√(x+(√(x+(√(............))))))))  ⇒y=(√(x+(√(x+(√(x+(√(x+......))))))))  ⇒y=(√(x+y  ))⇒ y^2 =x+y  ⇒2ydy=dx+dy ⇒ 2y(dy/dx) = 1+(dy/dx)  ⇒(2y−1)(dy/dx) = 1  ⇒  (dy/dx) = (1/(2y−1))    ((d/dx) (√(x+(√(x+(√(x+....∞))))))) = (1/(2(√(x+(√(x+(√(x+...∞))))))−1))..
$$\mathrm{Cherchons}\:\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$$$\mathrm{y}=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{…………}}}} \\ $$$$\Rightarrow\mathrm{y}=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+……}}}} \\ $$$$\Rightarrow\mathrm{y}=\sqrt{\mathrm{x}+\mathrm{y}\:\:}\Rightarrow\:\mathrm{y}^{\mathrm{2}} =\mathrm{x}+\mathrm{y} \\ $$$$\Rightarrow\mathrm{2ydy}=\mathrm{dx}+\mathrm{dy}\:\Rightarrow\:\mathrm{2y}\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{1}+\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$$$\Rightarrow\left(\mathrm{2y}−\mathrm{1}\right)\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{1}\:\:\Rightarrow\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{2y}−\mathrm{1}} \\ $$$$ \\ $$$$\left(\frac{\mathrm{d}}{\mathrm{dx}}\:\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+….\infty}}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+…\infty}}}−\mathrm{1}}.. \\ $$

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