3y² + x² - xy = 1\nfind \\frac{dy}{dx}\nchoose 1 answer\na \\frac{y - 2x}{6y - x}\nb \\frac{1 - 2x}{6y - 1}\nc

Explanation:

Step1: Differentiate both sides with respect to (x)

Differentiate (3y^{2}+x^{2}-xy = 1) term - by - term. Using the chain rule (\frac{d}{dx}(3y^{2})=3\times2y\frac{dy}{dx}=6y\frac{dy}{dx}), (\frac{d}{dx}(x^{2}) = 2x), and using the product rule (\frac{d}{dx}(xy)=x\frac{dy}{dx}+y). The derivative of the constant (1) is (0). So we have: (6y\frac{dy}{dx}+2x-(x\frac{dy}{dx}+y)=0)

Step2: Rearrange the terms to solve for (\frac{dy}{dx})

Expand the left - hand side: (6y\frac{dy}{dx}+2x - x\frac{dy}{dx}-y = 0) Group the terms with (\frac{dy}{dx}) together: ((6y - x)\frac{dy}{dx}+(2x - y)=0) Then ((6y - x)\frac{dy}{dx}=y - 2x)

Step3: Solve for (\frac{dy}{dx})

Divide both sides by (6y - x) (assuming (6y - x\neq0)): (\frac{dy}{dx}=\frac{y - 2x}{6y - x})

Answer:

A. (\frac{y - 2x}{6y - x})