Questions: If -y^3+x^3+x^2=-3y-2y^2 then find dy/dx at the point (-1,3).

Transcript text: If $-y^{3}+x^{3}+x^{2}=-3 y-2 y^{2}$ then find $\frac{d y}{d x}$ at the point $(-1,3)$.

Solution

Solution Steps

To find the derivative $\frac{dy}{dx}$ at the point $(-1, 3)$ for the given implicit equation, we will use implicit differentiation. This involves differentiating both sides of the equation with respect to $x$, applying the chain rule where necessary for terms involving $y$. After differentiating, we will solve for $\frac{dy}{dx}$ and substitute the point $(-1, 3)$ to find the specific value of the derivative at that point.

Step 1: Implicit Differentiation

We start with the equation given in the problem: \[ -y^3 + x^3 + x^2 = -3y - 2y^2. \] We rearrange it to: \[ x^3 + x^2 - y^3 + 2y^2 + 3y = 0. \] Next, we differentiate both sides with respect to $x$ using implicit differentiation. This gives us: \[ \frac{d}{dx}(x^3 + x^2 - y^3 + 2y^2 + 3y) = 0. \] Applying the chain rule, we find: \[ 3x^2 + 2x - 3y^2 \frac{dy}{dx} + 4y \frac{dy}{dx} + 3 \frac{dy}{dx} = 0. \]

Step 2: Solve for $\frac{dy}{dx}$

Rearranging the differentiated equation, we isolate $\frac{dy}{dx}$: \[ \frac{dy}{dx}(-3y^2 + 4y + 3) = - (3x^2 + 2x). \] Thus, we can express $\frac{dy}{dx}$ as: \[ \frac{dy}{dx} = \frac{- (3x^2 + 2x)}{-3y^2 + 4y + 3} = \frac{3x^2 + 2x}{3y^2 - 4y - 3}. \]

Step 3: Evaluate at the Point $(-1, 3)$

Now, we substitute the point $(-1, 3)$ into the expression for $\frac{dy}{dx}$: \[ \frac{dy}{dx} \bigg|_{(-1, 3)} = \frac{3(-1)^2 + 2(-1)}{3(3)^2 - 4(3) - 3}. \] Calculating the numerator: \[ 3(1) - 2 = 1, \] and the denominator: \[ 3(9) - 12 - 3 = 27 - 12 - 3 = 12. \] Thus, we have: \[ \frac{dy}{dx} \bigg|_{(-1, 3)} = \frac{1}{12}. \]