Question Number 205827 by mustafazaheen last updated on 31/Mar/24
$$\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} =\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{implceat}\:\mathrm{second}\:\mathrm{derivative} \\ $$
Answered by cortano12 last updated on 31/Mar/24
![3x^2 + 3y^2 y′ = 0 (dy/dx) = −(x^2 /y^2 )=−x^2 y^(−2) (d/dx) [ (dy/dx) ]= (d/dx) [−x^2 y^(−2) ] (d^2 y/dx^2 ) = −2xy^(−2) +2x^2 y^(−3) y′ (d^2 y/dx^2 ) = −((2x)/y^2 ) +((2x^2 )/y^3 ) . (((−x^2 )/y^2 )) = ((−2x)/y^2 )−((2x^4 )/y^5 ) =((−2xy^3 −2x^4 )/y^5 ) =((−2x(x^3 +y^3 ))/y^5 )= −((2x)/y^5 )](https://www.tinkutara.com/question/Q205830.png)
$$\:\mathrm{3}{x}^{\mathrm{2}} +\:\mathrm{3}{y}^{\mathrm{2}} {y}'\:=\:\mathrm{0} \\ $$$$\:\:\frac{{dy}}{{dx}}\:=\:−\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }=−{x}^{\mathrm{2}} {y}^{−\mathrm{2}} \\ $$$$\:\frac{{d}}{{dx}}\:\left[\:\frac{{dy}}{{dx}}\:\right]=\:\frac{{d}}{{dx}}\:\left[−{x}^{\mathrm{2}} {y}^{−\mathrm{2}} \:\right] \\ $$$$\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:−\mathrm{2}{xy}^{−\mathrm{2}} \:+\mathrm{2}{x}^{\mathrm{2}} {y}^{−\mathrm{3}} \:{y}'\: \\ $$$$\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:−\frac{\mathrm{2}{x}}{{y}^{\mathrm{2}} }\:+\frac{\mathrm{2}{x}^{\mathrm{2}} }{{y}^{\mathrm{3}} }\:.\:\left(\frac{−{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\frac{−\mathrm{2}{x}}{{y}^{\mathrm{2}} }−\frac{\mathrm{2}{x}^{\mathrm{4}} }{{y}^{\mathrm{5}} }\:=\frac{−\mathrm{2}{xy}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{4}} }{{y}^{\mathrm{5}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{−\mathrm{2}{x}\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)}{{y}^{\mathrm{5}} }=\:−\frac{\mathrm{2}{x}}{{y}^{\mathrm{5}} } \\ $$
Answered by TonyCWX08 last updated on 31/Mar/24
$$ \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$${f}\left({x},{y}\right)={x}^{\mathrm{3}} +{y}^{\mathrm{3}} −\mathrm{1} \\ $$$${f}_{{x}} \left({x},{y}\right)=\mathrm{3}{x}^{\mathrm{2}} \\ $$$${f}_{{y}} \left({x},{y}\right)=\mathrm{3}{y}^{\mathrm{2}} \\ $$$$ \\ $$$$\frac{{dy}}{{dx}} \\ $$$$=−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{3}{y}^{\mathrm{2}} } \\ $$$$=−\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} } \\ $$$$ \\ $$$${f}\left({x},{y}\right)=−\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} } \\ $$$${f}_{{x}} \left({x},{y}\right)=−\frac{\mathrm{2}{x}}{{y}^{\mathrm{2}} } \\ $$$${f}_{{y}} \left({x},{y}\right)=\frac{\mathrm{2}{x}^{\mathrm{2}} }{{y}^{\mathrm{3}} } \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} } \\ $$$$=−\frac{−\mathrm{2}{x}}{{y}^{\mathrm{2}} }\boldsymbol{\div}\frac{\mathrm{2}{x}^{\mathrm{2}} }{{y}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{2}{x}}{{y}^{\mathrm{2}} }×\frac{{y}^{\mathrm{3}} }{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$=\frac{{y}}{{x}} \\ $$
Commented by Frix last updated on 31/Mar/24
$$\mathrm{Obviously}\:\mathrm{this}\:\mathrm{is}\:\mathrm{wrong}. \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{1}\:\Leftrightarrow\:{y}=\left(\mathrm{1}−{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\Rightarrow \\ $$$${y}''=−\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}^{\mathrm{3}} \right)^{\frac{\mathrm{5}}{\mathrm{3}}} }=−\frac{\mathrm{2}{x}}{{y}^{\mathrm{5}} }\neq\frac{{y}}{{x}} \\ $$