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)

sec ф = 5/3 (quadrant 4)

let x = base

y = perp

z = hypotenuse

sec ф = hypotenuse/base = 5/3

⇒ z = 5 and x = 3

finding y:

z² = x² + y²

5² = 3² + y²

y² = 25-9

y² = 16

y = 4

now finding all other geometric functions of theta:

sin ф = y/z = 4/5

sin ф = -4/5 (sin is negative in the 4th quadrant)

cos ф = x/z = 3/5

cos ф = 3/5 (cos is positive in 4th quadrant)

tan ф = y/x = 4/3

tan ф = -4/3 (tan is negative in 4th quadrant)

cosec ф = z/y = 5/4

cosec ф = -5/4 (cosec is negative in the 4th quadrant)

cot ф = x/y = 3/4

tan ф = -3/4 (tan is negative in 4th quadrant)

the function makes a strait line horizontally with no x intercept but a y intercept at 9.