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We are tasked with finding \( y' \) for the implicit equation \( x^2 y^3 + x + 2y = -34 \) at the point \( (2, -2) \).

### Steps:
1. **Differentiate implicitly**: We need to take the derivative of the equation with respect to \( x \), treating \( y \) as a function of \( x \) (i.e., using the chain rule for terms involving \( y \)).

   The equation is:
   \[
   x^2 y^3 + x + 2y = -34
   \]

   Differentiating term by term:
   - For \( x^2 y^3 \), apply the product rule:
     \[
     \frac{d}{dx} \left( x^2 y^3 \right) = 2x y^3 + x^2 \cdot 3y^2 \cdot y'
     \]
   - For \( x \), the derivative is:
     \[
     \frac{d}{dx} (x) = 1
     \]
   - For \( 2y \), the derivative is:
     \[
     \frac{d}{dx} (2y) = 2y'
     \]
   - The right-hand side, \( -34 \), is a constant, so its derivative is zero.

2. **Collect all terms**:
   After differentiating, the equation becomes:
   \[
   2x y^3 + 3x^2 y^2 y' + 1 + 2y' = 0
   \]

3. **Solve for \( y' \)**:
   Isolate the terms with \( y' \):
   \[
   3x^2 y^2 y' + 2y' = - (2x y^3 + 1)
   \]
   Factor out \( y' \):
   \[
   y' (3x^2 y^2 + 2) = - (2x y^3 + 1)
   \]
   Solve for \( y' \):
   \[
   y' = \frac{-(2x y^3 + 1)}{3x^2 y^2 + 2}
   \]

4. **Substitute \( (x, y) = (2, -2) \)**:
   - \( x = 2 \)
   - \( y = -2 \)

   Now substitute these values into the equation for \( y' \):
   \[
   y' = \frac{-(2(2)(-2)^3 + 1)}{3(2)^2 (-2)^2 + 2}
   \]
   Simplify the numerator:
   \[
   2(2)(-8) + 1 = -32 + 1 = -31
   \]
   Simplify the denominator:
   \[
   3(4)(4) + 2 = 48 + 2 = 50
   \]
   So,
   \[
   y' = \frac{-(-31)}{50} = \frac{31}{50}
   \]

Thus, the value of \( y' \) at the point \( (2, -2) \) is \( \frac{31}{50} \).

Would you like more details or have any questions?

Here are 5 related questions you could explore:
1. How do you apply implicit differentiation to higher-order terms involving both \( x \) and \( y \)?
2. What is the geometric interpretation of \( y' \) for implicit curves?
3. How does the product rule apply to terms like \( x^2 y^3 \)?
4. What is the difference between implicit and explicit differentiation?
5. How can you check if your implicit differentiation result is correct?

**Tip:** When differentiating implicitly, always remember to apply the chain rule to terms involving \( y \) since \( y \) is a function of \( x \).

Find y' for x^2y^3 + x + 2y = -34 at the point (2, -2).

Learn how to apply implicit differentiation to solve for y' in the equation x^2y^3 + x + 2y = -34 at the point (2, -2). This step-by-step guide covers chain rule and product rule applications, suitable for Grades 11-12 and Calculus I.

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