If \(x^{3} + y^{3} = 28\), then the value of \(\displaystyle\frac{dy}{dx}\) at the point \((3, 1)\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-9\)
By implicit differentiation with respect to x:
\( \displaystyle 3x^{2} + 3y^{2}\frac{dy}{dx} = 0 \)
\( \displaystyle \frac{dy}{dx} = -\frac{x^{2}}{y^{2}} \)
At (3, 1), which lies on the curve since 27 + 1 = 28:
\( \displaystyle \frac{dy}{dx} = -\frac{9}{1} = -9 \)