xy^3-x^3y=2
I used implicit derivation to find (y')=(3x^2y-y^3)/(3xy^2-x^3)
2007-04-08 06:11:02 · 5 answers · asked by 8 3 in Science & Mathematics ➔ Mathematics
It is true. Congratulations!
3y^2 y' x -y' x^3 - 3x^2y-y^3=0
3y^2 y' x -y' x^3 = 3x^2y-y^3
y'=(3x^2y-y^3)/(3xy^2-x^3)
2007-04-08 06:21:45 · answer #1 · answered by iyiogrenci 6 · 1⤊ 0⤋
Yep! You're right!
xy^3 - x^3y = 2
[(x)(3y^2)(dy/dx) + (1)(y^3)] - [(3x^2)(y) + (x^3)(1)(dy/dx)] = 0
[(x)(3y^2)(dy/dx) + (y^3)] - [(3x^2)(y) + (x^3)(dy/dx)] = 0
(x)(3y^2)(dy/dx) + (y^3) - (3x^2)(y) - (x^3)(dy/dx) = 0
(y^3) - (3x^2)(y) = (x^3)(dy/dx) - (x)(3y^2)(dy/dx)
(y^3) - (3x^2)(y) = dy/dx (x^3 - 3xy^2)
dy/dx = (y^3 - 3x^2y) / (x^3 - 3xy^2)
If you want to go a little further....
dy/dx = y(y^2 - 3x^2) / x(x^2 - 3y^2)
2007-04-08 06:27:47 · answer #2 · answered by Anonymous · 1⤊ 0⤋
let's see...
y^3 + 3xy^2y' - 3x^2y + x^3y' = 0
y'(3xy^2 - x^3) = (3x^2y-y^3)/(3xy^2-x^3)\
survey says... DINGDINGDINGDINGDINGDING!!!
2007-04-08 06:22:21 · answer #3 · answered by ǝɔnɐs ǝɯosǝʍɐ Lazarus'd- DEI 6 · 0⤊ 0⤋
were you told to find the y' ??...
then if so.. count x as a constant.?
3xy^2 - (x^3y)(lnx)
oh no.. im not sure .
2007-04-08 06:21:04 · answer #4 · answered by Jami 3 · 0⤊ 1⤋
CONGRATULATIONS!!!!!!!!!YOU ARE CORRECT!!!!!!!
2007-04-08 06:42:42 · answer #5 · answered by Tarik 2 · 0⤊ 0⤋