Calculate the gradient of the curve \(x^{3} + y^{3} - 2xy = 11\) at (2, -1).

Assessment: WAEC SSCE - Further Mathematics - 2013 (Essay) Subject: Further Mathematics

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Gradient of \(x^{3}+y^{3}-2xy=11\) at \((2,-1)\).

Differentiate implicitly with respect to \(x\), using the product rule on \(2xy\):

\[3x^{2}+3y^{2}\frac{dy}{dx}-2\left(y+x\frac{dy}{dx}\right)=0\]\[3x^{2}+3y^{2}\frac{dy}{dx}-2y-2x\frac{dy}{dx}=0\]

Collect the derivative terms:

\[\frac{dy}{dx}\left(3y^{2}-2x\right)=2y-3x^{2}\]\[\frac{dy}{dx}=\frac{2y-3x^{2}}{3y^{2}-2x}\]

At \((2,-1)\):

\[\frac{dy}{dx}=\frac{2(-1)-3(2)^{2}}{3(-1)^{2}-2(2)}=\frac{-2-12}{3-4}=\frac{-14}{-1}=14\]

The gradient of the curve at \((2,-1)\) is \(14\).

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