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.