To solve this problem, we are asked to find the second derivative of $y = e^{x^3}$, which is $\frac{d^2y}{dx^2}$.

Step 1: First derivative $\frac{dy}{dx}$

We start by finding the first derivative of $y = e^{x^3}$. Using the chain rule:

$\frac{dy}{dx} = \frac{d}{dx} \left( e^{x^3} \right) = e^{x^3} \cdot \frac{d}{dx} (x^3) = e^{x^3} \cdot 3x^2$

So, the first derivative is:

$\frac{dy}{dx} = 3x^2 e^{x^3}$

Step 2: Second derivative $\frac{d^2y}{dx^2}$

Now, we need to differentiate $\frac{dy}{dx} = 3x^2 e^{x^3}$ again, using the product rule. The product rule states that:

$\frac{d}{dx} \left( f(x) g(x) \right) = f'(x) g(x) + f(x) g'(x)$

Let $f(x) = 3x^2$ and $g(x) = e^{x^3}$.

• $f'(x) = \frac{d}{dx} (3x^2) = 6x$
• $g'(x) = \frac{d}{dx} \left( e^{x^3} \right) = e^{x^3} \cdot 3x^2$

Now applying the product rule:

$\frac{d^2y}{dx^2} = 6x e^{x^3} + 3x^2 \cdot (3x^2 e^{x^3})$

Simplifying the second term:

$3x^2 \cdot 3x^2 = 9x^4$

Thus, the second derivative becomes:

$\frac{d^2y}{dx^2} = 6x e^{x^3} + 9x^4 e^{x^3}$

We can factor out $e^{x^3}$:

$\frac{d^2y}{dx^2} = \left( 6x + 9x^4 \right) e^{x^3}$

Final Answer:

The correct option is:

$\boxed{(6x + 9x^4)e^{x^3}}$

This corresponds to the first option in the list.

Would you like further details on this? Here are 5 related questions to deepen your understanding:

1. How do you apply the chain rule to exponential functions?
2. What is the product rule and when do you use it?
3. How can you extend this to higher-order derivatives?
4. What is the general approach for differentiating composite functions like $e^{x^n}$?
5. How do you factor out common terms after differentiation?

Tip: When applying the product and chain rules, breaking down each term separately can help avoid errors in longer derivatives.