To solve this, we need to find the derivatives of the function $f(x) = e^{e^x}$.

Step 1: First Derivative, $f'(x)$

1. Using the chain rule, the derivative of $f(x) = e^{e^x}$ with respect to $x$ is: $f'(x) = e^{e^x} \cdot \frac{d}{dx}(e^x)$
2. Since $\frac{d}{dx}(e^x) = e^x$, we get: $f'(x) = e^{e^x} \cdot e^x$
3. Thus, the first derivative is: $f'(x) = e^{e^x} \cdot e^x$

Step 2: Second Derivative, $f''(x)$

To find $f''(x)$, we differentiate $f'(x) = e^{e^x} \cdot e^x$ with respect to $x$.

1. Applying the product rule, we have: $f''(x) = \frac{d}{dx}\left(e^{e^x}\right) \cdot e^x + e^{e^x} \cdot \frac{d}{dx}(e^x)$
2. We already know $\frac{d}{dx}(e^x) = e^x$, and from the first derivative, $\frac{d}{dx}(e^{e^x}) = e^{e^x} \cdot e^x$. So we get: $f''(x) = \left(e^{e^x} \cdot e^x\right) \cdot e^x + e^{e^x} \cdot e^x$
3. Simplifying this, we obtain: $f''(x) = e^{e^x} \cdot e^{2x} + e^{e^x} \cdot e^x$
4. Factoring out $e^{e^x} \cdot e^x$, we get: $f''(x) = e^{e^x} \cdot e^x \cdot (e^x + 1)$

So, the second derivative is: $f''(x) = e^{e^x} \cdot e^x \cdot (e^x + 1)$

Let me know if you'd like more details on these steps or have any questions!

Related Questions

1. How would you find the third derivative of $f(x) = e^{e^x}$?
2. What is the general formula for the $n$-th derivative of $e^{e^x}$?
3. How does the chain rule apply when differentiating functions of functions?
4. How do exponential functions behave under differentiation?
5. What is the importance of the product rule in finding higher-order derivatives?

Tip

When differentiating complex functions, applying the chain rule and product rule correctly helps avoid mistakes, especially with functions within exponentials.