If \(3x^{2} + 2y^{2} + xy + x - 7 = 0\), find \(\frac{\mathrm d y}{\mathrm d x}\) at the point (-2, 1).
Question 1 Report
If \(3x^{2} + 2y^{2} + xy + x - 7 = 0\), find \(\frac{\mathrm d y}{\mathrm d x}\) at the point (-2, 1).
Answer Details
Differentiate \(3x^2 + 2y^2 + xy + x - 7 = 0\) implicitly with respect to \(x\). Use the product rule on \(xy\):
\[6x + 4y\frac{dy}{dx} + \left(y + x\frac{dy}{dx}\right) + 1 = 0.\]
Collect the \(\dfrac{dy}{dx}\) terms:
\[\frac{dy}{dx}(4y + x) = -(6x + y + 1) \;\Rightarrow\; \frac{dy}{dx} = -\frac{6x + y + 1}{4y + x}.\]
At the point \((-2, 1)\):
\[\frac{dy}{dx} = -\frac{6(-2) + 1 + 1}{4(1) + (-2)} = -\frac{-12 + 2}{4 - 2} = -\frac{-10}{2} = 5.\]
Answer: \(\dfrac{dy}{dx} = 5\) at \((-2, 1)\).
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