If \(3x^{2} + 2y^{2} + xy + x - 7 = 0\), find \(\frac{\mathrm d
FURTHER MATHEMATICS
WAEC 2010
If \(3x^{2} + 2y^{2} + xy + x - 7 = 0\), find \(\frac{\mathrm d y}{\mathrm d x}\) at the point (-2, 1).
Explanation
\(3x^{2} + 2y^{2} + xy + x - 7 = 0\)
We differentiate y implicitly with respect to x.
\(6x + 2(2y)\frac{\mathrm d y}{\mathrm d x} + y + x\frac{\mathrm d y}{\mathrm d x} + 1 = 0\)
\(\frac{\mathrm d y}{\mathrm d x} = \frac{- y - 1 - 6x}{4y + x}\)
= \(\frac{-(6x + y + 1)}{4y + x}\).
At the point (-2, 1),
\(\frac{\mathrm d y}{\mathrm d x} = \frac{-(-12 + 1 + 1)}{4(1) - 2}\)
= \(\frac{-(-10)}{2} = 5\)
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