Question

3y² + x² - xy = 1
find \frac{dy}{dx}
choose 1 answer
a \frac{y - 2x}{6y - x}
b \frac{1 - 2x}{6y - 1}
c

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 \(6y - x
eq0\)): \(\frac{dy}{dx}=\frac{y - 2x}{6y - x}\)

Answer:

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