If \(x^{2} - xy + 2y^{2} = 8\), then the value of \(\displaystyle\frac{dy}{dx}\) at the point \((2, 2)\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-\displaystyle\frac{1}{3}\)
By implicit differentiation, using the product rule on xy:
\( \displaystyle 2x - y - x\frac{dy}{dx} + 4y\frac{dy}{dx} = 0 \)
\( \displaystyle \frac{dy}{dx}(4y - x) = y - 2x \)
\( \displaystyle \frac{dy}{dx} = \frac{y - 2x}{4y - x} \)
At (2, 2), which lies on the curve since 4 − 4 + 8 = 8:
\( \displaystyle \frac{dy}{dx} = \frac{2 - 4}{8 - 2} = -\frac{2}{6} = -\frac{1}{3} \)