Let `f:R to R` be a function given by
`f(x+y)=f(x) f(y)"for all "x,y in R`
`"If "f(x)=1+xg(x),log_(e)2, "where "lim_(x to 0) g(x)=1. "Then, f'(x)=`

Options

A. `log_(e) 2^(f(x))`
B. `log_(e) (f(x))^(2)`
C. `log_(e)2`
D. none of these

Explanation

To solve the problem, we need to find the derivative \( f'(x) \) of the function \( f \) defined by the functional equation \( f(x+y) = f(x)f(y) \) and the specific form \( f(x) = 1 + xg(x) \log 2 \) where \( \lim_{x \to 0} g(x) = 1 \).

### Step-by-step Solution:

1. **Start with the functional equation**:
We know that \( f(x+y) = f(x)f(y) \) for all \( x, y \in \mathbb{R} \). This suggests that \( f \) is an exponential function.

2. **Differentiate \( f(x) \)**:
We can use the definition of the derivative:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

3. **Express \( f(x+h) \)**:
Using the functional equation, we can express \( f(x+h) \) as:
\[ f(x+h) = f(x)f(h) \]
Thus, we have:
\[ f'(x) = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} \]

4. **Factor out \( f(x) \)**:
We can factor \( f(x) \) out of the limit:
\[ f'(x) = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h} \]

5. **Substitute \( f(h) \)**:
We know from the given form that:
\[ f(h) = 1 + h g(h) \log 2 \]
Therefore:
\[ f(h) - 1 = h g(h) \log 2 \]

6. **Substitute into the limit**:
Now substituting this back into the limit gives:
\[ f'(x) = f(x) \lim_{h \to 0} \frac{h g(h) \log 2}{h} \]
The \( h \) cancels out:
\[ f'(x) = f(x) \lim_{h \to 0} g(h) \log 2 \]

7. **Evaluate the limit**:
Since \( \lim_{h \to 0} g(h) = 1 \), we have:
\[ f'(x) = f(x) \cdot 1 \cdot \log 2 = f(x) \log 2 \]

8. **Final expression**:
Therefore, we conclude that:
\[ f'(x) = f(x) \log 2 \]

### Final Answer:
\[ f'(x) = f(x) \log 2 \]