Questions: Consider the following equation. 4 x^3+3 x y-2 y^3=3 Find the slope of the tangent line at (1,-1).

Solution Steps

We start by differentiating the given equation implicitly with respect to \( x \): \[ 4x^3 + 3xy - 2y^3 = 3 \] Differentiating term by term: \[ \frac{d}{dx}(4x^3) + \frac{d}{dx}(3xy) - \frac{d}{dx}(2y^3) = \frac{d}{dx}(3) \] This gives: \[ 12x^2 + 3\left(y + x\frac{dy}{dx}\right) - 6y^2\frac{dy}{dx} = 0 \]

Rearrange the equation to solve for \(\frac{dy}{dx}\): \[ 12x^2 + 3y + 3x\frac{dy}{dx} - 6y^2\frac{dy}{dx} = 0 \] Group the terms involving \(\frac{dy}{dx}\): \[ 3x\frac{dy}{dx} - 6y^2\frac{dy}{dx} = -12x^2 - 3y \] Factor out \(\frac{dy}{dx}\): \[ \frac{dy}{dx}(3x - 6y^2) = -12x^2 - 3y \] Now, solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{-12x^2 - 3y}{3x - 6y^2} \]

Substitute \(x = 1\) and \(y = -1\) into the expression for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{-12(1)^2 - 3(-1)}{3(1) - 6(-1)^2} \] Simplify the numerator and denominator: \[ \frac{dy}{dx} = \frac{-12 + 3}{3 - 6} = \frac{-9}{-3} = 3 \]

Final Answer