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Find y'' with Implicit Differentiation for x^2+y^2=3y?
I'm able to find the first derivative of x^2+y^2=3y, which is -2x/(3-2y).
But I'm pretty confused about how to find y''. I get that you should come to a point where you can substitute in your original function and receive 0, but I don't know how to get there. Any help?
2 Answers
- 2 years agoFavorite Answer
y' = -2x / (3 - 2y)
y' = 2x / (2y - 3)
In general, I like to get rid of as many negative signs as possible
Now use the quotient rule
u = 2x
u' = 2
v = 2y - 3
v' = 2 * y' = 2 * 2x / (2y - 3) = 4x / (2y - 3)
y'' =>
(v * u' - u * v') / v^2 =>
(2 * (2y - 3) - 2x * 4x / (2y - 3)) / (2y - 3)^2 =>
(1/(2y - 3)) * (2 * (2y - 3)^2 - 8x^2) / (2y - 3)^2 =>
(2 * (2y - 3)^2 - 8x^2) / (2y - 3)^3 =>
2 * ((2y - 3)^2 - 4x^2) / (2y - 3)^3 =>
2 * (2y - 3 - 2x) * (2y - 3 + 2x) / (2y - 3)^3 =>
2 * (2y - 2x - 3) * (2y + 2x - 3) / (2y - 3)^3
x^2 + y^2 = 3y
2x * dx + 2y * dy = 3 * dy
2x * dx = 3 * dy - 2y * dy
2x / (3 - 2y) = dy/dx
That's the correct 1st derivative
y' = 2x / (3 - 2y)
u = 2x
u' = 2
v = 3 - 2y
v' = -2 * y' = -2 * 2x / (3 - 2y) = -4x / (3 - 2y)
y'' =>
((3 - 2y) * 2 - 2x * (-4x) / (3 - 2y)) / (3 - 2y)^2 =>
(1/(3 - 2y)) * (2 * (3 - 2y)^2 + 8x^2) / (3 - 2y)^2 =>
(2 * (3 - 2y)^2 + 8x^2) / (3 - 2y)^3 =>
(2 * (9 - 12y + 4y^2) + 8x^2) / (3 - 2y)^3 =>
(18 - 24y + 8y^2 + 8x^2) / (3 - 2y)^3 =>
(18 - 24y + 8 * (x^2 + y^2)) / (3 - 2y)^3 =>
(18 - 24y + 8 * 3y) / (3 - 2y)^3 =>
(18 - 24y + 24y) / (3 - 2y)^3 =>
18 / (3 - 2y)^3 =>
18 * (3 - 2y)^(-3) =>
-18 * (2y - 3)^(-3)
- Anonymous2 years ago
Answer
