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23 July, 02:55

The first derivative of x^2+2y^2=16 is - x / (2y), the second is - (2y^2+x^2) / (4y^3). Find the third implicit derivative of x^2+2y^2=16.

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  1. 23 July, 04:49
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    d³y/dx³ = (-2xy² - 3x³ - 4xy²) / (8y⁵)

    Step-by-step explanation:

    d²y/dx² = (-2y² - x²) / (4y³)

    Take the derivative (use quotient rule and chain rule):

    d³y/dx³ = [ (4y³) (-4y dy/dx - 2x) - (-2y² - x²) (12y² dy/dx) ] / (4y³) ²

    d³y/dx³ = [ (-16y⁴ dy/dx - 8xy³ - (-24y⁴ dy/dx - 12x²y² dy/dx) ] / (16y⁶)

    d³y/dx³ = (-16y⁴ dy/dx - 8xy³ + 24y⁴ dy/dx + 12x²y² dy/dx) / (16y⁶)

    d³y/dx³ = ((8y⁴ + 12x²y²) dy/dx - 8xy³) / (16y⁶)

    d³y/dx³ = ((2y² + 3x²) dy/dx - 2xy) / (4y⁴)

    Substitute:

    d³y/dx³ = ((2y² + 3x²) (-x / (2y)) - 2xy) / (4y⁴)

    d³y/dx³ = ((2y² + 3x²) (-x) - 4xy²) / (8y⁵)

    d³y/dx³ = (-2xy² - 3x³ - 4xy²) / (8y⁵)
  2. 23 July, 05:32
    0
    y''' = 3 (x² + 2y²) / (4y³)

    Step-by-step explanation:

    x² + 2y² = 16

    2x + 4y (y') = 0

    4y (y') = - 2x

    2y (y') = - x

    y' = - x/2y

    2y (y') = - 1

    2[y'*y' + y*y"] = - 1

    2 (y') ² + 2y (y") = - 1

    2 (-x/2y) ² + 2y (y") = - 1

    2 (x²/4y²) + 2y (y") = - 1

    2y (y") = - 1 - 2 (x²/4y²)

    8y³ (y") = - 4y² - 2x²

    4y³ (y") = - 2y² - x²

    y" = [-x² - 2y²] : (4y³)

    y" = - (x² + 2y²) / (4y³)

    4y³ (y") = - 2y² - x²

    4y³ (y"') + 12y² (y') (y") = - 4y (y') - 2x

    4y³ (y''') + 12y³[ - (x² + 2y²) / (4y³) ] =

    -4y (-x/2y) - 2x

    4y³ (y''') + 3 (-x²-2y²) = 2x - 2x

    4y³ (y''') = 3 (x² + 2y²)

    y''' = 3 (x² + 2y²) / (4y³)
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