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Question Number 145163 by mim24 last updated on 02/Jul/21

Answered by mathmax by abdo last updated on 02/Jul/21

$$\mathrm{y}=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+...+\mathrm{\infty }}}}\Rightarrow \mathrm{y}=\sqrt{\mathrm{x}+\mathrm{y}}\Rightarrow {\mathrm{y}}^2=\mathrm{x}+\mathrm{y}\Rightarrow$$$${\mathrm{y}}^2-\mathrm{y}-\mathrm{x}=0\to \mathrm{\Delta }=1+4\mathrm{x}\Rightarrow \mathrm{y}=\frac{1+\sqrt{1+4\mathrm{x}}}{2}\mathrm{or}\mathrm{y}=\frac{1-\sqrt{1+4\mathrm{x}}}{2}$$$$\mathrm{but}\mathrm{y}⩾0\Rightarrow \mathrm{y}=\frac{1+\sqrt{4\mathrm{x}+1}}{2}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\times \frac{4}{2\sqrt{4\mathrm{x}+1}}=\frac{1}{\sqrt{4\mathrm{x}+1}}\Rightarrow$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sqrt{4\mathrm{x}+1}}$$

Answered by puissant last updated on 02/Jul/21

$$\mathrm{Cherchons}\frac{\mathrm{dy}}{\mathrm{dx}}$$$$\mathrm{y}=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{............}}}}$$$$\Rightarrow \mathrm{y}=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+......}}}}$$$$\Rightarrow \mathrm{y}=\sqrt{\mathrm{x}+\mathrm{y}}\Rightarrow {\mathrm{y}}^2=\mathrm{x}+\mathrm{y}$$$$\Rightarrow 2\mathrm{ydy}=\mathrm{dx}+\mathrm{dy}\Rightarrow 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=1+\frac{\mathrm{dy}}{\mathrm{dx}}$$$$\Rightarrow (2\mathrm{y}-1)\frac{\mathrm{dy}}{\mathrm{dx}}=1\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2\mathrm{y}-1}$$$$$$$$\left(\frac{\mathrm{d}}{\mathrm{dx}}\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+....\mathrm{\infty }}}}\right)=\frac{1}{2\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+...\mathrm{\infty }}}}-1}..$$

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