what's the first derivative of x^2-2xy+3y^2=8?

Question

i meant dy/dx

Answer 1

2xdx-2ydx-2xdy+6ydy=0
then with a little algebra we have

dy/dx=(2x-2y)/(-2x+6y)=
dy/dx=(x-y)/(3y-x)

Answer 2

2

Answer 3

x^2 - 2xy + 3y^2 = 8

Your first step would be to implicitly differentiate with respect to x. Remember that the derivative of y is dy/dx.

Differentiating, we get

2x - 2[y + x(dy/dx)] + 6y(dy/dx) = 8

Expanding, we get

2x - 2y - x(dy/dx) + 6y(dy/dx) = 8

Now, bring everything that DOESN'T have a dy/dx to the right hand side.

-x(dy/dx) + 6y(dy/dx) = 8 - 2x + 2y

Factor out dy/dx, to get

dy/dx [-x + 6y] = (8 - 2x + 2y)

Now, divide both sides by [-x + 6y] to isolate dy/dx

dy/dx = (8 - 2x + 2y) / (-x + 6y)

Answer 4

Can you give more information? The first derivative of that equation with respect to x or y?

The first derivative of this equation with respect to x (dy/dx) is:
2x - 2y dx + 3y²dx = 0

The first derivative of this equation with respect to y is:
x²dy - 2x dy + 6y = 0

Answer 5

I suppose you want to solve for dy/dx.

The derivative of xy is x dy/dx + y * 1 by the product rule and implicit differentiation.

Also by implicit differentiation the derivative of y^2 is (2y) dy/dx.

2x - 2 (x dy/dx + y * 1) + 3 (2y) dy/dx = 0

2x - 2x dy/dx - 2y + 6y dy/dx = 0

Move all terms with dy/dx to the right side and factor out dy/dx

2x-2y=2x dy/dx - 6y dy/dx = dy/dx (2x-6y)

Divide and you get for dy/dx = (2x-2y)/(2x-6y)
=2(x-y) / (2(x-3y)) = (x-y) / (x-3y)

Answer 6

Let's differentiate both sides of this equation. 2x-2y-2xdy/dx+6ydy/dx=0. So 2(x-y)+(6y-2x)dy/dx=0. So 2(x-y)=(2x-6y)dy/dx so dy/dx=2(x-y)/(2x-6y) so dy/dx= (x-y)/(x-3y).

Answer 7

x^2-2xy+3y^2=8
dy/dx = y' = 2x -2xy' -2y +6yy'
y' + 2xy' -6yy' = 2x - 2y
y'(1+2x -6y) = 2(x - y)
y' = dy/dx = 2(x-y)/(1+2x-6y)

Answer 8

2xdx-2xdy-2ydx+6ydy=0