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)

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)

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)