Question Number 153965 by ZiYangLee last updated on 12/Sep/21
$$\mathrm{If}\:\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{xy}+{y}^{\mathrm{2}} =\mathrm{1},\:\mathrm{prove}\:\mathrm{that}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{\mathrm{2}}{\left({x}−{y}\right)^{\mathrm{3}} } \\ $$
Answered by ARUNG_Brandon_MBU last updated on 12/Sep/21
$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{xy}+{y}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{6}{x}−\mathrm{2}\left({x}\frac{{dy}}{{dx}}+{y}\right)+\mathrm{2}{y}\frac{{dy}}{{dx}}=\mathrm{0}\:\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{6}{x}−\mathrm{2}{y}}{\mathrm{2}\left({x}−{y}\right)}=\frac{\mathrm{3}{x}−{y}}{{x}−{y}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{\left({x}−{y}\right)\left(\mathrm{3}−\frac{{dy}}{{dx}}\right)−\left(\mathrm{3}{x}−{y}\right)\left(\mathrm{1}−\frac{{dy}}{{dx}}\right)}{\left({x}−{y}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:=\frac{\left({x}−{y}\right)\left(\mathrm{3}−\frac{\mathrm{3}{x}−{y}}{{x}−{y}}\right)−\left(\mathrm{3}{x}−{y}\right)\left(\mathrm{1}−\frac{\mathrm{3}{x}−{y}}{{x}−{y}}\right)}{\left({x}−{y}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:=\frac{−\mathrm{2}{y}−\left(\mathrm{3}{x}−{y}\right)\left(\frac{−\mathrm{2}{x}}{{x}−{y}}\right)}{\left({x}−{y}\right)^{\mathrm{2}} }=\frac{\left(−\mathrm{2}{xy}+\mathrm{2}{y}^{\mathrm{2}} \right)+\left(\mathrm{6}{x}^{\mathrm{2}} −\mathrm{2}{xy}\right)}{\left({x}−{y}\right)^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{4}{xy}}{\left({x}−{y}\right)^{\mathrm{3}} }=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{xy}\right)+\mathrm{6}{x}^{\mathrm{2}} −\mathrm{4}{xy}}{\left({x}−{y}\right)^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{\left({x}−{y}\right)^{\mathrm{3}} } \\ $$