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Question Number 153965 by ZiYangLee last updated on 12/Sep/21

If 3x^2 −2xy+y^2 =1, prove that (d^2 y/dx^2 )=(2/((x−y)^3 ))

$$\mathrm{If}\:\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{xy}+{y}^{\mathrm{2}} =\mathrm{1},\:\mathrm{prove}\:\mathrm{that}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{\mathrm{2}}{\left({x}−{y}\right)^{\mathrm{3}} } \\ $$

Answered by ARUNG_Brandon_MBU last updated on 12/Sep/21

3x^2 −2xy+y^2 =1 6x−2(x(dy/dx)+y)+2y(dy/dx)=0 ⇒(dy/dx)=((6x−2y)/(2(x−y)))=((3x−y)/(x−y)) (d^2 y/dx^2 )=(((x−y)(3−(dy/dx))−(3x−y)(1−(dy/dx)))/((x−y)^2 )) =(((x−y)(3−((3x−y)/(x−y)))−(3x−y)(1−((3x−y)/(x−y))))/((x−y)^2 )) =((−2y−(3x−y)(((−2x)/(x−y))))/((x−y)^2 ))=(((−2xy+2y^2 )+(6x^2 −2xy))/((x−y)^3 )) =((2y^2 +6x^2 −4xy)/((x−y)^3 ))=((2(1−3x^2 +2xy)+6x^2 −4xy)/((x−y)^3 )) =(2/((x−y)^3 ))

$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{xy}+{y}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{6}{x}−\mathrm{2}\left({x}\frac{{dy}}{{dx}}+{y}\right)+\mathrm{2}{y}\frac{{dy}}{{dx}}=\mathrm{0}\:\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{6}{x}−\mathrm{2}{y}}{\mathrm{2}\left({x}−{y}\right)}=\frac{\mathrm{3}{x}−{y}}{{x}−{y}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{\left({x}−{y}\right)\left(\mathrm{3}−\frac{{dy}}{{dx}}\right)−\left(\mathrm{3}{x}−{y}\right)\left(\mathrm{1}−\frac{{dy}}{{dx}}\right)}{\left({x}−{y}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:=\frac{\left({x}−{y}\right)\left(\mathrm{3}−\frac{\mathrm{3}{x}−{y}}{{x}−{y}}\right)−\left(\mathrm{3}{x}−{y}\right)\left(\mathrm{1}−\frac{\mathrm{3}{x}−{y}}{{x}−{y}}\right)}{\left({x}−{y}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:=\frac{−\mathrm{2}{y}−\left(\mathrm{3}{x}−{y}\right)\left(\frac{−\mathrm{2}{x}}{{x}−{y}}\right)}{\left({x}−{y}\right)^{\mathrm{2}} }=\frac{\left(−\mathrm{2}{xy}+\mathrm{2}{y}^{\mathrm{2}} \right)+\left(\mathrm{6}{x}^{\mathrm{2}} −\mathrm{2}{xy}\right)}{\left({x}−{y}\right)^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{4}{xy}}{\left({x}−{y}\right)^{\mathrm{3}} }=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{xy}\right)+\mathrm{6}{x}^{\mathrm{2}} −\mathrm{4}{xy}}{\left({x}−{y}\right)^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{\left({x}−{y}\right)^{\mathrm{3}} } \\ $$

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