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.

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)

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)