Let `f:R to R` be a function given by
`f(x+y)=f(x)f(y)"for all "x,yin R`
If `f(x) ne 0` for all `x in R` and f'(0) exists, then f'(x) equals

Options

A. f(x) for all `x in R`
B. f(x) f'(0) for all `x in R`
C. f(x)+f'(0) for all `x in R`
D. none of these

Explanation

To solve the problem, we need to analyze the functional equation given and derive the derivative of the function \( f \).

### Step-by-Step Solution:

1. **Understanding the Functional Equation**:
We are given the functional equation:
\[ f(x+y) = f(x)f(y) \quad \text{for all } x, y \in \mathbb{R} \]
This type of equation is characteristic of exponential functions.

2. **Finding \( f(0) \)**:
Let's set \( x = 0 \) and \( y = 0 \):
\[ f(0 + 0) = f(0)f(0) \implies f(0) = f(0)^2 \]
This implies that \( f(0)(1 - f(0)) = 0 \). Given that \( f(x) \neq 0 \) for all \( x \), we conclude:
\[ f(0) = 1 \]

3. **Differentiating the Function**:
We want to find \( f'(x) \). We start with the definition of the derivative:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Using the functional equation:
\[ f(x+h) = f(x)f(h) \]
We can substitute this into the derivative:
\[ f'(x) = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} = \lim_{h \to 0} \frac{f(x)(f(h) - 1)}{h} \]

4. **Taking the Limit**:
Since \( f(x) \) is a constant with respect to \( h \), we can factor it out:
\[ f'(x) = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h} \]
Let’s denote:
\[ L = \lim_{h \to 0} \frac{f(h) - 1}{h} \]
Then we have:
\[ f'(x) = f(x) L \]

5. **Finding \( L \)**:
We know that \( f'(0) \) exists. From our previous steps:
\[ f'(0) = L \]
Therefore, we can express \( f'(x) \) as:
\[ f'(x) = f(x) f'(0) \]

### Conclusion:
Thus, we have derived that:
\[ f'(x) = f(x) f'(0) \]