The given function is:

$f(x) = e^{x^3} - e^x$

We will compute the derivative $f'(x)$ step by step.

Step 1: Recall the derivative rules

1. The derivative of $e^u$, where $u = u(x)$, is given by: $\frac{d}{dx}e^u = e^u \cdot u'(x)$
2. The derivative of a sum/difference of functions is the sum/difference of their derivatives: $\frac{d}{dx}(f(x) \pm g(x)) = f'(x) \pm g'(x)$

Step 2: Differentiate the function

The function is: $f(x) = e^{x^3} - e^x$

Term 1: $e^{x^3}$

• Let $u = x^3$, so $u' = 3x^2$.
• Using the chain rule: $\frac{d}{dx}e^{x^3} = e^{x^3} \cdot 3x^2$

Term 2: $e^x$

• The derivative of $e^x$ is simply: $\frac{d}{dx}e^x = e^x$

Step 3: Combine the results

$f'(x) = \frac{d}{dx}e^{x^3} - \frac{d}{dx}e^x$ Substituting the derivatives: $f'(x) = e^{x^3} \cdot 3x^2 - e^x$

Final Answer:

$f'(x) = 3x^2 e^{x^3} - e^x$

Would you like me to verify this or provide a deeper explanation of any step?

Here are 5 follow-up questions:

1. How do you apply the chain rule to more complex functions?
2. What happens when $x = 0$? What is $f'(0)$?
3. Can we find the second derivative $f''(x)$?
4. What is the behavior of $f'(x)$ as $x \to \infty$?
5. How do exponential functions compare to polynomial growth in $f(x)$?

Tip: Always simplify inside-out when applying the chain rule to avoid confusion.