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22 January, 21:49

Find the first, second and third implicit derivatives of x^2+2y^2=16

I'm pretty sure the first derivative is - x / (2y), and the second is - (2y^2+x^2) / (4y^3)

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Answers (2)
  1. 22 January, 22:08
    0
    dy/dx = - x / (2y)

    d²y/dx² = (-2y² - x²) / (4y³)

    Step-by-step explanation:

    x² + 2y² = 16

    Take implicit derivative once:

    2x + 4y dy/dx = 0

    4y dy/dx = - 2x

    dy/dx = - x / (2y)

    Take derivative a second time:

    d²y/dx² = [ (2y) (-1) - (-x) (2 dy/dx) ] / (2y) ²

    d²y/dx² = (-2y + 2x dy/dx) / (4y²)

    d²y/dx² = (-2y + 2x (-x / (2y))) / (4y²)

    d²y/dx² = (-2y - x²/y) / (4y²)

    d²y/dx² = (-2y² - x²) / (4y³)
  2. 22 January, 22:51
    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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