Let's find the second derivative of the function $y = e^{2x^3}$.

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

The function is $y = e^{2x^3}$, and to differentiate it, we need to use the chain rule. The chain rule states that if $y = e^{u(x)}$, then $\frac{dy}{dx} = e^{u(x)} \cdot \frac{du}{dx}$.

Here, $u(x) = 2x^3$, so:

$\frac{du}{dx} = 6x^2$

Therefore, the first derivative is:

$\frac{dy}{dx} = e^{2x^3} \cdot 6x^2$

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

Now, to find the second derivative, we differentiate $\frac{dy}{dx} = 6x^2 e^{2x^3}$. We apply the product rule, which states that if $y = f(x) \cdot g(x)$, then:

$\frac{d}{dx} [f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x)$

In our case:

• $f(x) = 6x^2$, so $f'(x) = 12x$
• $g(x) = e^{2x^3}$, and from the first derivative, we know that $g'(x) = 6x^2 e^{2x^3}$

Thus, applying the product rule:

$\frac{d^2y}{dx^2} = (12x) \cdot e^{2x^3} + 6x^2 \cdot (6x^2 e^{2x^3})$

Simplifying this:

$\frac{d^2y}{dx^2} = 12x e^{2x^3} + 36x^4 e^{2x^3}$

Factor out the common term $e^{2x^3}$:

$\frac{d^2y}{dx^2} = e^{2x^3} (12x + 36x^4)$

Finally, factor the expression inside the parentheses:

$\frac{d^2y}{dx^2} = e^{2x^3} \cdot 12x (1 + 3x^3)$

Final Answer:

$\frac{d^2y}{dx^2} = 12x e^{2x^3} (1 + 3x^3)$

Would you like any further clarification or additional steps?

Here are some related questions:

1. How do you apply the chain rule to differentiate functions involving exponents?
2. What is the product rule in differentiation, and how does it work?
3. How can you use the chain rule in more complex exponential functions?
4. Can you explain how to factor expressions like $e^{2x^3} (12x + 36x^4)$?
5. What is the difference between the first and second derivatives in terms of their geometrical interpretation?

Tip: When taking higher-order derivatives, always look for common factors that can be factored out to simplify the expression.