x^2-5xy+3y^2=7
x^3 + y^3 = 8
2007-05-26 12:06:59 · 3 answers · asked by chetzel 3 in Science & Mathematics ➔ Mathematics
The answer has both veriables in it.
2007-05-26 12:21:26 · update #1
I hope you mean differentiate implicitly with respect to x.
1) 2x - 5y - 2x*dy/dx + 6(dy/dx) = 0
2x - 5y = dy/dx * (2x -6)
dy/dx = (2x - 5y)/(2x -6)
2) 3x^2 + 3y^2 * dy/dx = 0
3x^2 = -3y^2 * dy/dx
dy/dx = -x^2/y^2
2007-05-26 12:21:32 · answer #1 · answered by Lilovacookedrice 3 · 0⤊ 0⤋
ASSUMING you want dy/dx-type result,
d/dx (x^2) = 2x
d/dx (-5xy) = - 5y -5x dy/dx
d/dx (3y^2) = 6y dy/dx
d/dx( -7) = 0
Then 2x - 5y - 5x dy/dx + 6y dy/dx = 0
Pushing variables around (6y - 5x) dy/dx = 5y-2x
Divide both sides for the answer.
If I can do #1, you sure as hell can do #2.
2007-05-26 19:22:00 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
I would love to, but differentiate with respect to which variable?
2007-05-26 19:12:29 · answer #3 · answered by spiderdudejc31187 1 · 2⤊ 2⤋