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Question Number 117557 by abdullahquwatan last updated on 12/Oct/20
second derivative  x^2 +3y^2 =5
$$\mathrm{second}\:\mathrm{derivative} \\ $$$${x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} =\mathrm{5} \\ $$
Answered by Dwaipayan Shikari last updated on 12/Oct/20
2x+6y(dy/dx)=0⇒(dy/dx)=−(x/(3y))  2+6y(d^2 y/dx^2 )+6((dy/dx))^2 =0  2+6y(d^2 y/dx^2 )+((6x^2 )/(9y^2 )) =0  6y(d^2 y/dx^2 )=−((6x^2 +18y^2 )/(9y^2 ))  (d^2 y/dx^2 )=−((6x^2 +18y^2 )/(54y^3 ))=−((x^2 +3y^2 )/(9y^3 ))
$$\mathrm{2}{x}+\mathrm{6}{y}\frac{{dy}}{{dx}}=\mathrm{0}\Rightarrow\frac{{dy}}{{dx}}=−\frac{{x}}{\mathrm{3}{y}} \\ $$$$\mathrm{2}+\mathrm{6}{y}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{6}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2}+\mathrm{6}{y}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\frac{\mathrm{6}{x}^{\mathrm{2}} }{\mathrm{9}{y}^{\mathrm{2}} }\:=\mathrm{0} \\ $$$$\mathrm{6}{y}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=−\frac{\mathrm{6}{x}^{\mathrm{2}} +\mathrm{18}{y}^{\mathrm{2}} }{\mathrm{9}{y}^{\mathrm{2}} } \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=−\frac{\mathrm{6}{x}^{\mathrm{2}} +\mathrm{18}{y}^{\mathrm{2}} }{\mathrm{54}{y}^{\mathrm{3}} }=−\frac{{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} }{\mathrm{9}{y}^{\mathrm{3}} } \\ $$
Commented by abdullahquwatan last updated on 12/Oct/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by abdullahquwatan last updated on 12/Oct/20
sir 2+6y(d^2 y/dx^2 )+((6x^2 )/(9y^2 )) ?
$$\mathrm{sir}\:\mathrm{2}+\mathrm{6}{y}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\frac{\mathrm{6}{x}^{\mathrm{2}} }{\mathrm{9}{y}^{\mathrm{2}} }\:? \\ $$
Commented by abdullahquwatan last updated on 12/Oct/20
sir 2+6y(d^2 y/dx^2 )+((6x^2 )/(9y^2 )) ?
$$\mathrm{sir}\:\mathrm{2}+\mathrm{6}{y}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\frac{\mathrm{6}{x}^{\mathrm{2}} }{\mathrm{9}{y}^{\mathrm{2}} }\:? \\ $$
Commented by Dwaipayan Shikari last updated on 12/Oct/20
Oh yes! that was a mistake
$${Oh}\:{yes}!\:{that}\:{was}\:{a}\:{mistake} \\ $$
Commented by abdullahquwatan last updated on 12/Oct/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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