To find the second derivative of the implicit function $-2x^3 + y - y^2 = 0$ at the point $(-1, 2)$, we will follow the steps outlined:

Step 1: First Derivative using Implicit Differentiation

1. Differentiate both sides of the equation with respect to $x$:

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

   This gives:

   $-6x^2 + \frac{dy}{dx} - 2y \frac{dy}{dx} = 0$

2. Rearrange the equation to isolate $\frac{dy}{dx}$:

   $\frac{dy}{dx}(1 - 2y) = 6x^2$

3. Solve for $\frac{dy}{dx}$:

   $\frac{dy}{dx} = \frac{6x^2}{1 - 2y}$

Step 2: Evaluate the First Derivative at the Point $(-1, 2)$

Plugging in $x = -1$ and $y = 2$:

$\frac{dy}{dx} = \frac{6(-1)^2}{1 - 2(2)} = \frac{6}{1 - 4} = \frac{6}{-3} = -2$

Step 3: Second Derivative

1. Differentiate $\frac{dy}{dx} = \frac{6x^2}{1 - 2y}$ again with respect to $x$:

   Using the quotient rule:

   $\frac{d^2y}{dx^2} = \frac{(1 - 2y)\frac{d}{dx}(6x^2) - 6x^2\frac{d}{dx}(1 - 2y)}{(1 - 2y)^2}$

   Calculating each part:

   • $\frac{d}{dx}(6x^2) = 12x$
   • $\frac{d}{dx}(1 - 2y) = -2\frac{dy}{dx}$

   Substitute:

   $\frac{d^2y}{dx^2} = \frac{(1 - 2y)(12x) - 6x^2(-2\frac{dy}{dx})}{(1 - 2y)^2}$

2. Substitute $x = -1$, $y = 2$, and $\frac{dy}{dx} = -2$:

   Calculate $1 - 2(2) = 1 - 4 = -3$:

   $\frac{d^2y}{dx^2} = \frac{(-3)(12(-1)) - 6(-1)^2(-2)}{(-3)^2}$

   This simplifies to:

   $\frac{d^2y}{dx^2} = \frac{36 - 12}{9} = \frac{24}{9} = \frac{8}{3}$

Final Result

The second derivative at the point $(-1, 2)$ is

$\frac{d^2y}{dx^2} = \frac{8}{3}.$

Summary

1. The first derivative $\frac{dy}{dx}$ at the point $(-1, 2)$ is $-2$.
2. The second derivative $\frac{d^2y}{dx^2}$ at the point $(-1, 2)$ is $\frac{8}{3}$.

Would you like to go into more detail on any of these steps or have any other questions? Here are some related questions:

1. What is the geometric interpretation of the first derivative?
2. How does the second derivative relate to the concavity of the function?
3. Can you provide examples of other implicit functions and their derivatives?
4. What is the importance of the chain rule in implicit differentiation?
5. How would the second derivative change if the equation were different?

Tip: Always double-check your algebra when working with derivatives, as small errors can lead to incorrect results.