Let's go through the steps to find $f'(0)$ based on the image provided:

1. Given Function: $f'(x) = e^x \cdot g'(x) + g(x) \cdot e^x$

2. At $x = 0$: $f'(0) = e^0 \cdot g'(0) + g(0) \cdot e^0$

3. Simplify $e^0$: Since $e^0 = 1$, this simplifies the equation to: $f'(0) = 1 \cdot g'(0) + g(0) \cdot 1$ $f'(0) = g'(0) + g(0)$

4. Substitute Given Values: From the image, we are given: $g(0) = 4 \quad \text{and} \quad g'(0) = 2$ Therefore: $f'(0) = 2 + 4 = 6$

Final Answer:

$f'(0) = 6$

Would you like further explanation on any part of this? Here are some related questions:

1. What is the significance of $e^x$ in this function?
2. How does the product rule apply in this derivative?
3. Why does $e^0$ simplify to 1?
4. What other methods could be used to evaluate $f'(0)$?
5. How would this change if we were asked to find $f''(0)$?

Tip: Always remember that the derivative of $e^x$ with respect to $x$ is still $e^x$, which is a powerful property in calculus!