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

i know y' is 0 but is y'' 0.5 or -0.5?

4 Answers

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  • 1 month ago
    Favourite answer

    x² + 2xy + 3y² = 2

    when y = 1, 

    x² + 2x*1 + 3*1² = 2

    x² + 2x + 1 = 0

    x = -1

    x² + 2xy + 3y² = 2

    2x + 2( xy’ + y ) + 6y y’ = 0

    2x + 2xy’ + 2y + 6yy’ = 0

    when y = 1, x = -1

    2(-1) + 2( -y’ + 1 ) + 6y’ = 0

    -2 - 2y’ + 2 + 6y’ = 0

    y’ = 0

    ━━━

    2x + 2xy’ + 2y + 6yy’ = 0

    2 + 2( xy’’ + y’ ) + 2y’ + 6( yy’’ + y’y’) = 0

    2 + 2xy’’ + 2y’ + 2y’ + 6yy’’ + 6y’y’ = 0

    2 + 2xy’’ + 4y’ + 6yy’’ + 6y’y’ = 0

    when y = 1, x = -1, y’ = 0

    2 + 2(-1)y’’ + 4(0) + 6(1)y’’ + 6(0)(0) = 0

    y’’ = -½

    ━━━━

  • Philip
    Lv 6
    1 month ago

    x^2 +2xy +3y^2 = 2...(1). Notation : ID = ''implicit differentiation of''.;

    ID(1) gives 2x + 2y + 2xy' + 6yy' = 0, ie., (x+3y)y' +(x+y) = 0...(2).;

    ID(2) gives (x+3y)y'' + (1+3y')y' + 1+ y' = 0, ie., (x+3y)y'' + (2+3y')y' +1 = 0...(3).;

    When y = 1, (1) becomes x^2 +2x +1 = 0, ie., (x+1)^2 = 0 and x = -1.;

    For (x,y) = (-1,1), (2) becomes (-1+3)y' +(-1+1) = 0, ie., 2y' = 0, ie., y' = 0.;

    For (x,y) = (-1,1), (3) becomes (-1+3)y'' + 1 = 0, ie., 2y'' = -1, ie., y'' = -(1/2).

  • 1 month ago

    Why y = 1:

    x² + 2x(1) + 3(1)² = 2

    x² + 2x + 1 = 0

    (x + 1)² = 0

    x = -1

    x² + 2xy + 3y² = 2

    (d/dx)(x² + 2xy + 3y²) = (d/dx)2

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

    2xy' + 6yy' = -2x - 2y

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

    When y = 1, and x = -1:

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

    y' = 0

    y" = (d/dx)[-(x + y) / (x + 3y)]

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

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

    When y = 1, x = -11 and y' = 0:

    y" = -{[(-1) + 3(1)][1 + (0)] - [(-1) + (1)][1 + 3(0)] / [(-1) + 3(1)]²}

    y" = -(2 - 0) / 4

    y" = -1/2

    Hence, when y = 1, y' = 0 and y" = -1/2

  • 1 month ago

    Take derivatives with respect to x on both sides:

        2x + (2y + 2xy') + 6yy' = 0

        (2x + 6y) y' = -2x - 2y

        y' = - (x + y) / (x + 3y)

    The value of x matters, so solve for what it must be if y = 1:

        x^2 + 2x + 3 = 2    .... plug in y=1

        x^2 + 2x + 1 = 0

        (x + 1)^2 = 0

        x = -1

    So yes, y' = 0 because (x + y) in the numerator is (-1 + 1) = 0.  Take derivatives with respect to x again in that first equation above:

        2 + 2y' + (2xy'' + 2y') + 6[(y')^2 + yy''] = 0

        2 + 0  - 2y'' + 0 + 6(0^2 + y'') = 0

         2 - 2y'' + 6y'' = 0

        4y'' = -2

        y'' = -1/2

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