Let `f(x+y)=f(x)dotf(y)` for all `xa n dydot` Suppose `f(5)=2a n df^(prime)(0)=3.` Find `f^(prime)(5)dot`

Explanation

To solve the problem, we will follow the steps outlined in the video transcript and provide a detailed step-by-step solution.

### Step-by-Step Solution:

1. **Given Condition**: We have the functional equation \( f(x+y) = f(x) \cdot f(y) \) for all \( x \) and \( y \). We also know that \( f(5) = 2 \) and \( f'(0) = 3 \). We need to find \( f'(5) \).

2. **Using the Definition of Derivative**: The derivative of \( f \) at point \( x \) is defined as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

3. **Substituting \( x = 5 \)**: We substitute \( x = 5 \) into the derivative definition:
\[ f'(5) = \lim_{h \to 0} \frac{f(5+h) - f(5)}{h} \]

4. **Using the Functional Equation**: According to the given condition, we can express \( f(5+h) \) as:
\[ f(5+h) = f(5) \cdot f(h) \]
Thus, we can rewrite the derivative:
\[ f'(5) = \lim_{h \to 0} \frac{f(5) \cdot f(h) - f(5)}{h} \]

5. **Factoring Out \( f(5) \)**: We can factor out \( f(5) \) from the limit:
\[ f'(5) = f(5) \cdot \lim_{h \to 0} \frac{f(h) - 1}{h} \]

6. **Finding \( f(0) \)**: To find \( f(0) \), we use the functional equation with \( x = 5 \) and \( y = 0 \):
\[ f(5+0) = f(5) \cdot f(0) \implies f(5) = f(5) \cdot f(0) \]
Since \( f(5) \neq 0 \), we can divide both sides by \( f(5) \):
\[ 1 = f(0) \implies f(0) = 1 \]

7. **Finding \( f'(0) \)**: We know that \( f'(0) = 3 \). Thus:
\[ f'(5) = f(5) \cdot f'(0) = 2 \cdot 3 \]

8. **Final Calculation**: Therefore, we find:
\[ f'(5) = 6 \]

### Conclusion:
The value of \( f'(5) \) is \( 6 \).