Let's start by solving problem 11, which states:

Problem 11: Let $f(x) = e^{g(x)}$, where $g(0) = 6$ and $g'(0) = 2$. Find $f'(0)$.

Solution:

To find $f'(x)$, we need to differentiate $f(x) = e^{g(x)}$ using the chain rule.

1. The derivative of $f(x) = e^{g(x)}$ is: $f'(x) = e^{g(x)} \cdot g'(x)$
2. Now, substitute $x = 0$: $f'(0) = e^{g(0)} \cdot g'(0)$
3. We're given that $g(0) = 6$ and $g'(0) = 2$, so: $f'(0) = e^6 \cdot 2$ $f'(0) = 2e^6$

Thus, the value of $f'(0)$ is $2e^6$.

Would you like more details or any further explanation on this solution?

Here are five related questions for further practice:

1. If $h(x) = \ln(g(x))$ and $g(0) = 6$, $g'(0) = 2$, find $h'(0)$.
2. Differentiate $f(x) = e^{3g(x)}$ with respect to $x$.
3. Find the second derivative $f''(0)$ for the same function $f(x) = e^{g(x)}$.
4. Solve $f'(x)$ if $f(x) = e^{g(x)} + x^2g(x)$.
5. If $g'(x) = \sin(x)$, find $f'(x)$ for $f(x) = e^{g(x)}$.

Tip: Always apply the chain rule when differentiating composite functions like $e^{g(x)}$.