find dy/dx
x^2+y^2-xy=2
2007-10-22 05:21:29 · 4 answers · asked by tsLow 1 in Science & Mathematics ➔ Mathematics
ill just use y' instead of dy/dx
2x+2yy'-(xy'+y)=0
2x+2yy'-xy'-y=0
y'(2y-x)=y-2x
y'=(y-2x)/(2y-x)
2007-10-22 05:24:06 · answer #1 · answered by biowolf89 3 · 0⤊ 0⤋
Jeezzzz Biowolf. It's certainly took you long enough to do a simple derivative âº
Doug
2007-10-22 12:26:35 · answer #2 · answered by doug_donaghue 7 · 0⤊ 0⤋
here on differentiating both sides with respect to x we get,
2x+2y(dy/dx)-(x*dy/dx + y) =0
or, 2x+2y (dy/dx)-x*(dy/dx)-y=0
or,(2y-x)(dy/dx)= y-2x
or,dy/dx = (2y-x)/(y-2x)
this is the answer.
2007-10-22 12:31:45 · answer #3 · answered by sureswinter100 2 · 0⤊ 1⤋
2x + 2y (dy/dx) - (y + x dy/dx) = 0
(2y - x)(dy/dx) = y - 2x
dy/dx = (y - 2x) / (2y - x)
2007-10-22 14:35:00 · answer #4 · answered by Como 7 · 0⤊ 0⤋