If \(x^{2} + xy + y^{2} = 7\), then the value of \(\displaystyle\frac{dy}{dx}\) at the point \((1, 2)\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-\frac{4}{5}\)
The point lies on the curve, since \(1 + 2 + 4 = 7\). By implicit differentiation, with the product rule for the term xy:
\( \displaystyle 2x + \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0 \)
\( \displaystyle (x + 2y)\frac{dy}{dx} = -(2x + y) \)
\( \displaystyle \frac{dy}{dx} = -\frac{2x + y}{x + 2y} \)
At (1, 2):
\( \displaystyle \frac{dy}{dx} = -\frac{2 + 2}{1 + 4} \)
\( \displaystyle = -\frac{4}{5} \)