What is derivative of x^2+y^2-3xy=1 ?

The answer to derivative of x^2+y^2-3xy=1 is (dy)/(dx)=(-2x+3y)/(2y-3x)

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derivative of x^2+y^2-3xy=1

derivative of

x^{2} + y^{2} - 3 x y = 1

\frac{d y}{d x} = \frac{- 2 x + 3 y}{2 y - 3 x}

Solution steps

x^{2} + y^{2} - 3 x y = 1

Treat

y

y (x)

Differentiate both sides:

2 x + 2 y \frac{d y}{d x} - 3 (y + x \frac{d y}{d x}) = 0

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{- 2 x + 3 y}{2 y - 3 x}

\frac{d y}{d x} = \frac{- 2 x + 3 y}{2 y - 3 x}

7 x^{4} - 4 y = 4

4 x + 5 y = lo g_{10} (4 x - 3 y)

x^{4} + y^{4} = x^{2} y^{2}

x^{y} = y^{2}

y^{2} = - 4 x + 3

What is derivative of x^2+y^2-3xy=1 ?
The answer to derivative of x^2+y^2-3xy=1 is (dy)/(dx)=(-2x+3y)/(2y-3x)