Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

If (x^2)+2xy+(3y^2)=2, find y' and y'' when y=1?

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  • Vaman
    Lv 7
    2 months ago

    If (x^2)+2xy+(3y^2)=2, find y' and y'' when y=1? LET us find the x value. x^2 +2x +3=2, x^2 +2x+1=0 (x+1)^2= x=-1.

    Take the derivative.

    2x +2y +2x y'+ 6y y'=0,  y' ( 2x+6y)+2(x+y)=0 Put the values.

    y' 4=0, y'=Differentiate again,

    y''(2x+6y) + y'(2+6y')+2(1+y')=0. Put the values. y''(-2+6)+0+2=0

    y''= -2/4=1/2.

  • 2 months ago

    (x^2)+2xy+(3y^2)=2

    2x + 2(xy' + y) +6y*y' = 0

    2x + 2y + ( 2x + 6y )*y' = 0

    y' = -2( x + y )/2( x + 3y ) = - (x +y)/(x + 3y)

    y'(1) = ?

    when y =1

    ⇒ x² + 2x + 3 = 2

    ⇒ (x + 1)² = 0 ⇒ x = -1

    y'(1) = 0

    y" = -[ (x + 3y)(1 + y') - (x + y)(1 + 3y')]/(x + 3y)² 

    y"(1) = -[(2*1 -0)/4]

            = -1/2

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