The equation provided is:

\[
4x^2y = -4
\]

We are tasked with finding \(\frac{dy}{dx}\), which involves differentiating both sides of the equation with respect to \(x\). We'll use implicit differentiation since \(y\) is a function of \(x\).

### Step-by-step process:

1. **Differentiate both sides of the equation**:

   \[
   \frac{d}{dx}(4x^2y) = \frac{d}{dx}(-4)
   \]

2. **Use the product rule** for the left-hand side, since \(4x^2y\) is a product of two functions: \(4x^2\) and \(y\).

   The product rule is:
   \[
   \frac{d}{dx}(uv) = u'v + uv'
   \]
   Here, \(u = 4x^2\) and \(v = y\).

   Differentiating \(u = 4x^2\):
   \[
   \frac{du}{dx} = 8x
   \]
   
   So, applying the product rule:
   \[
   8xy + 4x^2\frac{dy}{dx}
   \]

3. **Differentiate the right-hand side**:

   Since \(-4\) is a constant, its derivative is 0:
   \[
   \frac{d}{dx}(-4) = 0
   \]

4. **Set up the equation**:

   After differentiating both sides, we get:
   \[
   8xy + 4x^2\frac{dy}{dx} = 0
   \]

5. **Solve for \(\frac{dy}{dx}\)**:

   Subtract \(8xy\) from both sides:
   \[
   4x^2\frac{dy}{dx} = -8xy
   \]

   Now divide both sides by \(4x^2\) (assuming \(x \neq 0\)):
   \[
   \frac{dy}{dx} = \frac{-8xy}{4x^2}
   \]

   Simplifying the right-hand side:
   \[
   \frac{dy}{dx} = \frac{-2y}{x}
   \]

Thus, the derivative is:

\[
\frac{dy}{dx} = \frac{-2y}{x}
\]

---

Would you like more details on any of the steps? Here are 5 related questions to consider:

1. How do you apply implicit differentiation to more complex equations?
2. Can you explain the product rule in more detail?
3. What happens if \(x = 0\) in this equation?
4. How do you solve implicit differentiation problems with higher powers?
5. Can implicit differentiation be used with trigonometric functions?

**Tip:** Always simplify your equations as much as possible after differentiating, it helps avoid unnecessary complexity later on!

Learn how to find dy/dx using implicit differentiation for the equation 4x^2y = -4. This problem explores the product rule and derivative techniques. Suitable for advanced calculus students in Grades 11-12.

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