A maths問題

1. 2x^3-(2x^2+1)y^2=3

6x^2 - [4xy^2 + (2x^2+1) 2y dy/dx] = 0

6x^2 - 4xy^2 - (2x^2+1) 2y dy/dx = 0

(2x^2+1) 2y dy/dx = 6x^2 - 4xy^2

dy/dx = (3x^2 - 2xy^2) / y(2x^2 +1)



2. (x^2+y^2)^4 = 2y

4(x^2+y^2)^3 (2x + 2y dy/dx) = 2 dy/dx

4x(x^2+y^2)^3 + 4y(x^2+y^2)^3 dy/dx = dy/dx

[4y(x^2+y^2)^3 - 1] dy/dx = -4x(x^2+y^2)^3

dy/dx = 4x(x^2+y^2)^3 / [1- 4y(x^2+y^2)^3]



请选我做最佳回答

1. 2x^3-(2x^2+1)y^2=3
6x^2 - [4xy^2 + (2x^2+1) 2y dy/dx] = 0
6x^2 - 4xy^2 - (2x^2+1) 2y dy/dx = 0
(2x^2+1) 2y dy/dx = 6x^2 - 4xy^2
dy/dx = (3x^2 - 2xy^2) / y(2x^2 +1)

2. (x^2+y^2)^4 = 2y
4(x^2+y^2)^3 (2x + 2y dy/dx) = 2 dy/dx
4x(x^2+y^2)^3 + 4y(x^2+y^2)^3 dy/dx = dy/dx
[4y(x^2+y^2)^3 - 1] dy/dx = -4x(x^2+y^2)^3
dy/dx = 4x(x^2+y^2)^3 / [1- 4y(x^2+y^2)^3]

2008-01-18 23:41:03 補充：
上面嗰位仁兄d 2x^3 = 6x^2

2x^3 - (2x^2+1)y^2 = 3
differentiate both sides w.r.t. x
6x - (2x^2+1)2y(dy/dx) + (y^2)(4x) = 0
-2y(2x^2+1)(dy/dx) = -6x-4xy^2
y(2x^2+1)(dy/dx) = 3x+2xy^2
dy/dx = [x(3+2y^2)]/[y(2x^2+1)]

(x^2+y^2)^4=2y
Differentiate both sides w.r.t. x
4[(x^2+y^2)^3](2x+2ydy/dx)=2dy/dx
8x(x^2+y^2)^3 + 8y(x^2+y^2)^3 dy/dx = 2dy/dx
4x(x^2+y^2)^3 = 1-4y(x^2+y^2)^3 dy/dx
dy/dx = 4x(x^2+y^2)^3 / [1-4y(x^2+y^2)^3]