✔ 最佳答案
x = √[(y-2)/(y+2)]
x^2 = (y-2)/(y+2)
x^2 = 1 - 4/(y+2)
dx^2/dx = d1/dx - d4/(y+2)dx
2x = 0 + 4/(y+2)^2 * dy/dx
x/2 = 1/(y+2)^2 * dy/dx
dy/dx = x(y+2)^2 /2
dy/dx = √[(y-2)/(y+2)] * (y+2)^2 / 2
dy/dx = (y-2)^(1/2) * (y+2)^(-1/2) * (y+2)^2 (0.5)
dy/dx = 0.5 (y02)^(1/2) (y+2)^(2 - 1/2)
d/ydx = 0.5 (y02)^(1/2) (y+2)^(3/2)
2011-05-30 18:44:04 補充：
Correction:
dy/dx = (y-2)^(1/2) * (y+2)^(-1/2) * (y+2)^2 (0.5)
dy/dx = 0.5 (y-2)^(1/2) (y+2)^(2 - 1/2)
d/ydx = 0.5 (y-2)^(1/2) (y+2)^(3/2)
2011-05-30 20:21:08 補充：
If you want to find dy/dx directly, u need to change the subject term to y,
x = √[(y-2)/(y+2)]
x^2 = (y-2)/(y+2)
x^2 y + 2x^2 = y - 2
2(x^2 + 1) = y (1-x^2)
y = 2(1+x^2)/(1-x^2)
2011-05-30 20:22:38 補充：
dy/dx = [d2(1+x^2)/dx (1-x^2) - d(1-x^2)/dx *2(1+x^2)]/(1-x^2)^2
= [4x(1-x^2) + 4x(1+x^2)]/(1-x^2)^2
= 4x(1-x^2+1+x^2)/(1-x^2)^2
= 4x/(1-x^2)^2
2011-05-30 20:26:11 補充：
= 4x(1-x^2+1+x^2)/(1-x^2)^2
= 8x/(1-x^2)^2
2011-05-30 20:28:01 補充：
= 8√[(y-2)/(y+2)] / ( 1 - (y-2)/(y+2))^2
= 8√[(y-2)/(y+2)]/[(y+2-y+2)^2 /(y+2)^2]
= 8√[(y-2)/(y+2)]/[16/(y+2)^2]
= 0.5 (y-2)^(1/2) (y+2)^(2 - 1/2)
= 0.5 (y-2)^(1/2) (y+2)^(3/2)
參考: Hope the solution can help you^^”
