https://file.aiquickdraw.com/mathbot/file/1728192889155n3rlbeb7.jpg

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!

We are to find f'(0) based on the derivative function f'(x) = e^x * g'(x) + g(x) * e^x, using given values g(0) = 4 and g'(0) = 2.

This problem involves finding f'(0) from a function that includes both a derivative term and an exponential term. Given g(0) = 4 and g'(0) = 2, the solution applies the product rule and properties of the exponential function.

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