Calculate the gradient of the curve \(x^{3} + y^{3} - 2xy = 11\) at (2, -1).
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Correct Answer: Option n
Explanation:
\(x^{3} + y^{3} - 2xy = 11\)
Differentiating implicitly,
\(3x^{2} + 3y^{2} \frac{\mathrm d y}{\mathrm d x} - 2y - 2x \frac{\mathrm d y}{\mathrm d x} = 0\)
\((3y^{2} - 2x) \frac{\mathrm d y}{\mathrm d x} = 2y - 3x^{2}\)
\(\frac{\mathrm d y}{\mathrm d x} = \frac{2y - 3x^{2}}{3y^{2} - 2x}\)
At (2, -1) , Gradient = \(\frac{2(-1) - 3(2^{2})}{3(-1)^{2} - 2(2)}\)
= \(\frac{-2 - 12}{3 - 4}\)
= \(\frac{-14}{-1} = 14\)
\(x^{3} + y^{3} - 2xy = 11\)
Differentiating implicitly,
\(3x^{2} + 3y^{2} \frac{\mathrm d y}{\mathrm d x} - 2y - 2x \frac{\mathrm d y}{\mathrm d x} = 0\)
\((3y^{2} - 2x) \frac{\mathrm d y}{\mathrm d x} = 2y - 3x^{2}\)
\(\frac{\mathrm d y}{\mathrm d x} = \frac{2y - 3x^{2}}{3y^{2} - 2x}\)
At (2, -1) , Gradient = \(\frac{2(-1) - 3(2^{2})}{3(-1)^{2} - 2(2)}\)
= \(\frac{-2 - 12}{3 - 4}\)
= \(\frac{-14}{-1} = 14\)