To find the second derivative of \( (2x^2 + 1)^3 \), we proceed step by step.

### Step 1: First derivative

Using the chain rule, let \( u = 2x^2 + 1 \), so the function becomes \( u^3 \). Then:
\[
\frac{d}{dx} \left( u^3 \right) = 3u^2 \cdot \frac{du}{dx}
\]

Now, we need to find \( \frac{du}{dx} \), where \( u = 2x^2 + 1 \):
\[
\frac{du}{dx} = 4x
\]

Substitute \( \frac{du}{dx} \) back into the derivative of \( u^3 \):
\[
\frac{d}{dx} \left( (2x^2 + 1)^3 \right) = 3(2x^2 + 1)^2 \cdot 4x = 12x(2x^2 + 1)^2
\]

### Step 2: Second derivative

Now, we differentiate \( 12x(2x^2 + 1)^2 \) using the product rule:
\[
\frac{d}{dx} \left( 12x(2x^2 + 1)^2 \right) = 12 \cdot \frac{d}{dx} \left( x(2x^2 + 1)^2 \right)
\]

Using the product rule:
\[
\frac{d}{dx} \left( x(2x^2 + 1)^2 \right) = (2x^2 + 1)^2 + x \cdot 2(2x^2 + 1) \cdot \frac{d}{dx}(2x^2 + 1)
\]

We already know \( \frac{d}{dx}(2x^2 + 1) = 4x \), so the second term becomes:
\[
x \cdot 2(2x^2 + 1) \cdot 4x = 8x^2(2x^2 + 1)
\]

Now, the derivative becomes:
\[
\frac{d}{dx} \left( x(2x^2 + 1)^2 \right) = (2x^2 + 1)^2 + 8x^2(2x^2 + 1)
\]

Substituting this into the expression for the second derivative:
\[
12 \cdot \left( (2x^2 + 1)^2 + 8x^2(2x^2 + 1) \right)
\]

Thus, the second derivative is:
\[
12(2x^2 + 1)^2 + 96x^2(2x^2 + 1)
\]

Would you like any more details or have any questions?

Here are some related questions:
1. How do you apply the chain rule in differentiation?
2. What are some applications of the second derivative in real-world problems?
3. How do you differentiate composite functions?
4. What is the significance of the product rule in differentiation?
5. How can higher-order derivatives help in analyzing function behavior?

**Tip:** The second derivative can provide information about the concavity and points of inflection of a function.

Learn how to compute the second derivative of (2x^2 + 1)^3 using the chain rule and product rule. This detailed step-by-step solution covers key concepts in differentiation, including higher-order derivatives. Suitable for Grades 11-12 or early college students.

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