Question Number 11552 by Nayon last updated on 28/Mar/17
$${find}\:\frac{{d}}{{dx}}\left({y}\right)\:{where}\:\:{y}=\overset{\sqrt{{x}}} {\:}\sqrt{\sqrt{{x}}} \\ $$$$ \\ $$
Answered by ajfour last updated on 28/Mar/17
$${y}\:=\:\left(\sqrt{{x}}\right)^{\frac{\mathrm{1}}{\sqrt{{x}}}} \\ $$$$\mathrm{ln}\:{y}\:=\frac{\mathrm{ln}\:\sqrt{{x}}}{\sqrt{{x}}} \\ $$$$\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dx}}\:=\:\left(\frac{\left(\frac{\mathrm{1}}{\sqrt{{x}}}\right)\:\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\right)\sqrt{{x}}\:−\frac{\mathrm{ln}\:\sqrt{{x}}}{\mathrm{2}\sqrt{{x}}}}{{x}}\right) \\ $$$$\frac{{dy}}{{dx}}\:=\:{y}\left(\frac{\mathrm{1}−\mathrm{ln}\:\sqrt{{x}}}{\mathrm{2}{x}\sqrt{{x}}}\right) \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{\left(\sqrt{{x}}\:\right)^{\frac{\mathrm{1}}{\sqrt{{x}}}} \:\:\left(\mathrm{1}−\mathrm{ln}\:\sqrt{{x}}\:\right)}{\mathrm{2}{x}\sqrt{{x}}}\:. \\ $$