find dy/dx x^2+y^2 =2xy?

Answer 1

dy/dx x^2+y^2 =2xy

lets do dy first then, dx second, since you have both x and y in the function

dy will be

2y dy = 2x

dx will be

2x dx =2y

you solve for dy and dx,
dy = 2x / 2y or x / y

dx = 2y / 2x or y / x

so dy/dx will be x/y/y/x
or x^2 / y^2

Answer 2

x^2+y^2=2xy
differenciate both sides wrt x;
2x=2y dy/dx=2y+2x dy/dx
collecting terms containing dy/dx;
2(y-x)dy/dx=2(y-x)
if x is different from y;
dy/dx=1
Or x^2+y^2=2xy
(y-x)^2=0
y-x=0
y=x
dy/dx=1

Answer 3

(x-y)^2=0
so 2(x-y)*(1-y´)=0 so y´=1 or x=y which gives y´=1

Answer 4

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

Answer 5

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