Questions: Differentiate implicitly to find the slope of the curve at the given point. y^3+y x^2+x^2-3 y^2=0 ;(-1,1) A. -1/2 B. -1 C. -2 D. 3/2

Solution Steps

To find the slope of the curve at the given point using implicit differentiation, we need to differentiate both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\). After differentiating, solve for \(\frac{dy}{dx}\) and substitute the given point \((-1, 1)\) to find the slope.

We start with the equation of the curve given by

\[ y^{3} + y x^{2} + x^{2} - 3 y^{2} = 0. \]

To find the slope \(\frac{dy}{dx}\), we differentiate both sides with respect to \(x\):

\[ \frac{d}{dx}(y^{3}) + \frac{d}{dx}(y x^{2}) + \frac{d}{dx}(x^{2}) - \frac{d}{dx}(3 y^{2}) = 0. \]

Using the product rule and chain rule, we obtain:

\[ 3y^{2} \frac{dy}{dx} + (x^{2} \frac{dy}{dx} + 2xy) + 2x - 6y \frac{dy}{dx} = 0. \]

Rearranging the equation gives:

\[ (3y^{2} - 6y + x^{2}) \frac{dy}{dx} + 2xy + 2x = 0. \]

Solving for \(\frac{dy}{dx}\) yields:

\[ \frac{dy}{dx} = -\frac{2x(y + 1)}{x^{2} + 3y^{2} - 6y}. \]

Now, we substitute the point \((-1, 1)\) into the expression for the slope:

\[ \frac{dy}{dx} = -\frac{2(-1)(1 + 1)}{(-1)^{2} + 3(1)^{2} - 6(1)}. \]

Calculating the numerator:

\[ -2(-1)(2) = 4, \]

and the denominator:

\[ 1 + 3 - 6 = -2. \]

Thus, we have:

\[ \frac{dy}{dx} = -\frac{4}{-2} = 2. \]

Final Answer