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Question Number 32510 by mondodotto@gmail.com last updated on 26/Mar/18

Commented by abdo imad last updated on 26/Mar/18

we have y =(√(1+(√(1+(√x) )))) ⇒y^2 −1 =(1+(√x) )^(1/2) ⇒2y y^, =(1/2)( (1/(2(√x))))^(−(1/2)) = (1/(2(√(2 ))(√(√x)))) ⇒ y^′ = (1/(4(√2)(√((√x) )) y)) = (1/(4(√2))) (^4 (√x))^(−1) ((√(1+(√(1+(√x))))) )^(−1) .

$${we}\:{have}\:{y}\:=\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{{x}}\:\:}}\:\:\Rightarrow{y}^{\mathrm{2}} \:−\mathrm{1}\:=\left(\mathrm{1}+\sqrt{{x}}\:\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow\mathrm{2}{y}\:{y}^{,} \:=\frac{\mathrm{1}}{\mathrm{2}}\left(\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}\:}\sqrt{\sqrt{{x}}}}\:\:\Rightarrow \\ $$$${y}^{'} \:\:=\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}\sqrt{\sqrt{{x}}\:}\:{y}}\:\:\:=\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\:\left(^{\mathrm{4}} \sqrt{{x}}\right)^{−\mathrm{1}} \:\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{{x}}}}\:\right)^{−\mathrm{1}} . \\ $$

Answered by Joel578 last updated on 26/Mar/18

(1) y^2 = 1 + (√(1 + (√x))) (y^2 − 1)^2 = 1 + (√x) 2(y^2 − 1)(2y)((dy/dx)) = (1/(2(√x))) (4y^3 − 4y)((dy/dx)) = (1/(2(√x))) (dy/dx) = (1/(2(√x)(4y^3 − 4y))) = (1/(8y(√x)(y^2 − 1)))

$$\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} \:=\:\mathrm{1}\:+\:\sqrt{\mathrm{1}\:+\:\sqrt{{x}}} \\ $$$$\left({y}^{\mathrm{2}} \:−\:\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{1}\:+\:\sqrt{{x}} \\ $$$$\mathrm{2}\left({y}^{\mathrm{2}} \:−\:\mathrm{1}\right)\left(\mathrm{2}{y}\right)\left(\frac{{dy}}{{dx}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}} \\ $$$$\left(\mathrm{4}{y}^{\mathrm{3}} \:−\:\mathrm{4}{y}\right)\left(\frac{{dy}}{{dx}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}\left(\mathrm{4}{y}^{\mathrm{3}} \:−\:\mathrm{4}{y}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{8}{y}\sqrt{{x}}\left({y}^{\mathrm{2}} \:−\:\mathrm{1}\right)} \\ $$

Answered by MJS last updated on 27/Mar/18

1. f(x)=(√x); f′(x)=(1/(2f(x))) (f(1+f(1+f(x))))′= =(1/(2f(x)))×(1/(2f(1+f(x))))×(1/(2f(1+f(1+f(x)))))= (1/(8(√x)×(√(1+(√x)))×(√(1+(√(1+(√x))))))) 2a. hyperbola (y^2 /4)−(x−1)^2 =1 2b. hyperbola ((10y^2 )/9)−(x^2 /9)=1 2c. circle (x^2 /(25))+(((y+5)^2 )/(25))=1

$$\mathrm{1}. \\ $$$${f}\left({x}\right)=\sqrt{{x}};\:{f}'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}{f}\left({x}\right)} \\ $$$$\left({f}\left(\mathrm{1}+{f}\left(\mathrm{1}+{f}\left({x}\right)\right)\right)\right)'= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{f}\left({x}\right)}×\frac{\mathrm{1}}{\mathrm{2}{f}\left(\mathrm{1}+{f}\left({x}\right)\right)}×\frac{\mathrm{1}}{\mathrm{2}{f}\left(\mathrm{1}+{f}\left(\mathrm{1}+{f}\left({x}\right)\right)\right)}= \\ $$$$\frac{\mathrm{1}}{\mathrm{8}\sqrt{{x}}×\sqrt{\mathrm{1}+\sqrt{{x}}}×\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{{x}}}}} \\ $$$$ \\ $$$$\mathrm{2a}.\:\mathrm{hyperbola} \\ $$$$\frac{{y}^{\mathrm{2}} }{\mathrm{4}}−\left({x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2b}.\:\mathrm{hyperbola} \\ $$$$\frac{\mathrm{10}{y}^{\mathrm{2}} }{\mathrm{9}}−\frac{{x}^{\mathrm{2}} }{\mathrm{9}}=\mathrm{1} \\ $$$$\mathrm{2c}.\:\mathrm{circle} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{25}}+\frac{\left({y}+\mathrm{5}\right)^{\mathrm{2}} }{\mathrm{25}}=\mathrm{1} \\ $$

Commented by byaw last updated on 27/Mar/18

2c. it is a circle centre (0,−5) and radius 5 units.

$$\mathrm{2}{c}.\:{it}\:{is}\:{a}\:{circle}\:{centre}\:\left(\mathrm{0},−\mathrm{5}\right)\:{and} \\ $$$$\:\:\:\:\:\:\:\:{radius}\:\mathrm{5}\:{units}. \\ $$

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