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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

$$wehavey=\sqrt{1+\sqrt{1+\sqrt{x}}}\Rightarrow y^2-1={(1+\sqrt{x})}^{\frac{1}{2}}$$$$\Rightarrow 2y{y}^{,}=\frac{1}{2}{\left(\frac{1}{2\sqrt{x}}\right)}^{-\frac{1}{2}}=\frac{1}{2\sqrt{2}\sqrt{\sqrt{x}}}\Rightarrow$$$$y^{{}^{\prime }}=\frac{1}{4\sqrt{2}\sqrt{\sqrt{x}}y}=\frac{1}{4\sqrt{2}}{\left({}^4\sqrt{x}\right)}^{-1}{\left(\sqrt{1+\sqrt{1+\sqrt{x}}}\right)}^{-1}.$$

Answered by Joel578 last updated on 26/Mar/18

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

Answered by MJS last updated on 27/Mar/18

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

Commented by byaw last updated on 27/Mar/18

$$2c.itisacirclecentre(0,-5)and$$$$radius5units.$$

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