Question Number 42463 by abdo.msup.com last updated on 26/Aug/18
$${let}\:{y}\:\:=\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\mathrm{2}}}} \\ $$$${calculate}\:\:\frac{{dy}}{{dx}} \\ $$
Commented by maxmathsup by imad last updated on 26/Aug/18
$${we}\:{have}\:{y}^{\mathrm{2}} \:={x}+\sqrt{{x}+\sqrt{{x}+\mathrm{2}}\:}\:\:\Rightarrow\left({y}^{\mathrm{2}} −{x}\right)^{\mathrm{2}} \:={x}+\sqrt{{x}+\mathrm{2}}\:\Rightarrow \\ $$$${y}^{\mathrm{4}} −\mathrm{2}{xy}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \:={x}+\sqrt{{x}+\mathrm{2}}\:\:\:{let}\:{derivate}\:\Rightarrow \\ $$$$\mathrm{4}{y}^{'} \:{y}^{\mathrm{3}} \:−\mathrm{2}\left({y}^{\mathrm{2}} \:+\mathrm{2}{xyy}'\right)\:+\mathrm{2}{x}\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{2}}}\:\Rightarrow \\ $$$$\left(\mathrm{4}{y}^{\mathrm{3}} \:−\mathrm{4}{xy}\right){y}^{'} \:−\mathrm{2}{y}^{\mathrm{2}} \:+\mathrm{2}{x}\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{2}}}\:\Rightarrow \\ $$$$\left(\mathrm{4}{y}^{\mathrm{3}} −\mathrm{4}{xy}\right){y}^{'} \:=\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{2}}}\:\Rightarrow \\ $$$${y}^{'} \:=\frac{\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{2}}}}{\mathrm{4}{y}^{\mathrm{3}} −\mathrm{4}{xy}}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 26/Aug/18
$${y}^{\mathrm{2}} ={x}+\sqrt{{x}+\sqrt{{x}+\mathrm{2}}\:} \\ $$$$\mathrm{2}{y}\frac{{dy}}{{dx}}=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\sqrt{{x}+\mathrm{2}}}}\right)×\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{2}}}\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\left(\sqrt{{x}+\sqrt{{x}+\mathrm{2}}}\right.}\right\}×\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{2}}}\right)}{\mathrm{2}{y}} \\ $$