Question Number 32510 by mondodotto@gmail.com last updated on 26/Mar/18
Commented by abdo imad last updated on 26/Mar/18
$${we}\:{have}\:{y}\:=\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{{x}}\:\:}}\:\:\Rightarrow{y}^{\mathrm{2}} \:−\mathrm{1}\:=\left(\mathrm{1}+\sqrt{{x}}\:\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow\mathrm{2}{y}\:{y}^{,} \:=\frac{\mathrm{1}}{\mathrm{2}}\left(\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}\:}\sqrt{\sqrt{{x}}}}\:\:\Rightarrow \\ $$$${y}^{'} \:\:=\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}\sqrt{\sqrt{{x}}\:}\:{y}}\:\:\:=\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\:\left(^{\mathrm{4}} \sqrt{{x}}\right)^{−\mathrm{1}} \:\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{{x}}}}\:\right)^{−\mathrm{1}} . \\ $$
Answered by Joel578 last updated on 26/Mar/18
$$\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} \:=\:\mathrm{1}\:+\:\sqrt{\mathrm{1}\:+\:\sqrt{{x}}} \\ $$$$\left({y}^{\mathrm{2}} \:−\:\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{1}\:+\:\sqrt{{x}} \\ $$$$\mathrm{2}\left({y}^{\mathrm{2}} \:−\:\mathrm{1}\right)\left(\mathrm{2}{y}\right)\left(\frac{{dy}}{{dx}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}} \\ $$$$\left(\mathrm{4}{y}^{\mathrm{3}} \:−\:\mathrm{4}{y}\right)\left(\frac{{dy}}{{dx}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}\left(\mathrm{4}{y}^{\mathrm{3}} \:−\:\mathrm{4}{y}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{8}{y}\sqrt{{x}}\left({y}^{\mathrm{2}} \:−\:\mathrm{1}\right)} \\ $$
Answered by MJS last updated on 27/Mar/18
$$\mathrm{1}. \\ $$$${f}\left({x}\right)=\sqrt{{x}};\:{f}'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}{f}\left({x}\right)} \\ $$$$\left({f}\left(\mathrm{1}+{f}\left(\mathrm{1}+{f}\left({x}\right)\right)\right)\right)'= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{f}\left({x}\right)}×\frac{\mathrm{1}}{\mathrm{2}{f}\left(\mathrm{1}+{f}\left({x}\right)\right)}×\frac{\mathrm{1}}{\mathrm{2}{f}\left(\mathrm{1}+{f}\left(\mathrm{1}+{f}\left({x}\right)\right)\right)}= \\ $$$$\frac{\mathrm{1}}{\mathrm{8}\sqrt{{x}}×\sqrt{\mathrm{1}+\sqrt{{x}}}×\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{{x}}}}} \\ $$$$ \\ $$$$\mathrm{2a}.\:\mathrm{hyperbola} \\ $$$$\frac{{y}^{\mathrm{2}} }{\mathrm{4}}−\left({x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2b}.\:\mathrm{hyperbola} \\ $$$$\frac{\mathrm{10}{y}^{\mathrm{2}} }{\mathrm{9}}−\frac{{x}^{\mathrm{2}} }{\mathrm{9}}=\mathrm{1} \\ $$$$\mathrm{2c}.\:\mathrm{circle} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{25}}+\frac{\left({y}+\mathrm{5}\right)^{\mathrm{2}} }{\mathrm{25}}=\mathrm{1} \\ $$
Commented by byaw last updated on 27/Mar/18
$$\mathrm{2}{c}.\:{it}\:{is}\:{a}\:{circle}\:{centre}\:\left(\mathrm{0},−\mathrm{5}\right)\:{and} \\ $$$$\:\:\:\:\:\:\:\:{radius}\:\mathrm{5}\:{units}. \\ $$