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 + xfrac{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)