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The problem asks to use implicit differentiation to find \( y' \) for the equation:

\[
x^2 y - 4x^3 + 16 = 0
\]

and then evaluate \( y' \) at the point \( (2,4) \).

### Step 1: Implicit Differentiation
Differentiate both sides of the equation with respect to \( x \), remembering to apply the product rule when differentiating \( x^2 y \) and the chain rule when differentiating terms involving \( y \):

\[
\frac{d}{dx}(x^2 y) - \frac{d}{dx}(4x^3) + \frac{d}{dx}(16) = 0
\]

- For \( x^2 y \), use the product rule: 
  \[
  \frac{d}{dx}(x^2 y) = 2x y + x^2 \frac{dy}{dx} = 2xy + x^2 y'
  \]

- For \( -4x^3 \), differentiate normally: 
  \[
  \frac{d}{dx}(-4x^3) = -12x^2
  \]

- \( 16 \) is a constant, so its derivative is 0.

Now the differentiated equation is:

\[
2xy + x^2 y' - 12x^2 = 0
\]

### Step 2: Solve for \( y' \)
To isolate \( y' \), rearrange the equation:

\[
x^2 y' = 12x^2 - 2xy
\]

Now divide both sides by \( x^2 \):

\[
y' = 12 - \frac{2xy}{x^2}
\]

Simplify the fraction:

\[
y' = 12 - \frac{2y}{x}
\]

### Step 3: Evaluate \( y' \) at \( (x, y) = (2, 4) \)

Substitute \( x = 2 \) and \( y = 4 \) into the equation:

\[
y' = 12 - \frac{2(4)}{2} = 12 - 4 = 8
\]

So, \( y' \big|_{(2,4)} = 8 \).

### Final Answer:
- \( y' = 12 - \frac{2y}{x} \)
- \( y'(2,4) = 8 \)

Do you want further details or explanations? Here are five related questions:

1. What is the product rule and when do we use it in differentiation?
2. How does implicit differentiation differ from explicit differentiation?
3. What are some real-world applications of implicit differentiation?
4. How can we verify if a point satisfies an implicitly defined curve?
5. Can implicit differentiation be used for second derivatives? How?

**Tip:** Implicit differentiation is especially useful when \( y \) cannot be easily solved in terms of \( x \).

Use implicit differentiation to find y', and then evaluate y' for x^2y - 4x^3 + 16 = 0 at the point (2,4).

Learn how to use implicit differentiation to find the derivative y' for the equation x^2y - 4x^3 + 16 = 0, and how to evaluate y' at the point (2,4). This problem covers key calculus concepts, including the product rule and the chain rule, suitable for students in Grades 11-12.

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