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Let's solve the given equation step by step.

### Step 1: Solve the equation \( y(x^2 + 16) = 32 \) for \( y \).

\[
y(x^2 + 16) = 32
\]

Divide both sides by \( x^2 + 16 \) to express \( y \) explicitly:

\[
y = \frac{32}{x^2 + 16}
\]

### Step 2: Differentiate \( y = \frac{32}{x^2 + 16} \) with respect to \( x \).

We will use the **quotient rule** for differentiation. Recall that:

\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}
\]

For \( y = \frac{32}{x^2 + 16} \):
- \( u = 32 \), so \( u' = 0 \).
- \( v = x^2 + 16 \), so \( v' = 2x \).

Applying the quotient rule:

\[
\frac{dy}{dx} = \frac{0 \cdot (x^2 + 16) - 32 \cdot 2x}{(x^2 + 16)^2}
\]

Simplify:

\[
\frac{dy}{dx} = \frac{-64x}{(x^2 + 16)^2}
\]

### Final Answer:

\[
\frac{dy}{dx} = \frac{-64x}{(x^2 + 16)^2}
\]

Let me know if you'd like a deeper explanation or further clarifications!

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Here are some related questions to explore further:
1. How do you apply the product rule for differentiation?
2. What are some typical applications of the quotient rule?
3. How does the derivative behave as \( x \) approaches 0 or infinity in this function?
4. What is the meaning of the negative sign in the derivative result?
5. How would you find the second derivative of this function?

**Tip:** In many calculus problems involving fractions, the quotient rule is useful, but sometimes rewriting the function as a product of terms with negative exponents can simplify the differentiation.

Solve the equation y(x^2 + 16) = 32 for y to find an explicit expression for y and then calculate dy/dx.

Learn how to solve the equation y(x^2 + 16) = 32 for y and calculate dy/dx using the quotient rule. This solution covers key calculus concepts, ideal for students in Grades 11-12.

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