Question

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)

Answer 1

y''' = 3(x² + 2y²)/(4y³)

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)

Answer 2

dy/dx = -x/(2y)

d²y/dx² = (-2y² − x²) / (4y³)

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)