Let `f(x+y)=f(x)f(y)` for all x and y. Then what is f'(5) equal to [where f' (x)] is the derivative of f(x)] ?

Options

A. `f(5)f'(0)`
B. `f(5)-f'(0)`
C. `f(5)f(0)`
D. `f(5)+f'(0)`

Answer

To solve the problem, we start with the functional equation given:

1. **Given**: \( f(x+y) = f(x)f(y) \) for all \( x \) and \( y \).

2. **Step 1**: Let's find \( f(0) \). Set \( x = 0 \) and \( y = 0 \):
\[ f(0 + 0) = f(0)f(0) \implies f(0) = f(0)^2 \]
This implies \( f(0)(1 - f(0)) = 0 \). Thus, \( f(0) = 0 \) or \( f(0) = 1 \). Since \( f(x+y) = f(x)f(y) \) must hold for all \( x \) and \( y \), we discard \( f(0) = 0 \) because it would make \( f(x) = 0 \) for all \( x \). Therefore, we have:
\[ f(0) = 1 \]

3. **Step 2**: Differentiate the functional equation. We differentiate both sides with respect to \( y \):
\[ \frac{d}{dy} f(x+y) = \frac{d}{dy} (f(x)f(y)) \]
Using the chain rule on the left side and the product rule on the right side:
\[ f'(x+y) = f(x)f'(y) \]

4. **Step 3**: Set \( y = 0 \) in the differentiated equation:
\[ f'(x+0) = f(x)f'(0) \implies f'(x) = f(x)f'(0) \]

5. **Step 4**: This shows that the derivative \( f'(x) \) is proportional to \( f(x) \). We can express this as:
\[ f'(x) = k f(x) \]
where \( k = f'(0) \).

6. **Step 5**: We can solve this differential equation. The general solution to \( f'(x) = k f(x) \) is:
\[ f(x) = Ce^{kx} \]
where \( C \) is a constant.

7. **Step 6**: Using the condition \( f(0) = 1 \):
\[ f(0) = Ce^{k \cdot 0} = C = 1 \]
Thus, we have:
\[ f(x) = e^{kx} \]

8. **Step 7**: Now, we need to find \( f'(5) \):
\[ f'(x) = k e^{kx} \]
Therefore,
\[ f'(5) = k e^{5k} \]

9. **Step 8**: Since we don't have the specific value of \( k \), we conclude that \( f'(5) \) is expressed as:
\[ f'(5) = k e^{5k} \]

### Final Answer:
The value of \( f'(5) \) is \( k e^{5k} \).