use implicit differentiation to find dy/dx if 2xy=y^2?

Question

thank you for the homework help!

please show the algebra when solving

Answer 1

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

Answer 2

Okay, your first step will be to differentiate both sides. So, by the product rule (1st * der. 2nd + 2nd * der. 1st, or xy'+yx') and power rule (der. x^n=nx^(n-1))

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

Remember that y^1 is just y. There's a dy/dx because we took the derivative of y.

Now, get the dy/dx on the same side. Subtract 2x(dy/dx) from each side.

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

Rearrange it to 2y=dy/dx(2y-2x)

Solve for dy-dx! (divide by (2y-2x))

2y/(2y-2x)=dy/dx.

Simplify! We can take a two out of both parts, so

dy/dx=y/(y-x)

Answer 3

2x(dy/dx)+2y=2y(dy/dx)
Solve for dy/dx = y/(y-x)

Answer 4

2xy=y^2
2xy'+2y =2yy'
xy' -yy' = - y
y' = -y/x-y)