Let f(x) be a function satisfying `f(x+y)=f(x)f(y)` for all `x,y in R` and f(x)=1+xg(x) where `underset(x to 0)lim g(x)=1`. Then f'(x) is equal to

Options

A. `f(x)`
B. `[f(x)]^(2)`
C. `g(x)`
D. None of these

Answer

To solve the problem step by step, let's start with the given conditions and derive the required result.

### Step 1: Understanding the Functional Equation
We are given that \( f(x+y) = f(x)f(y) \) for all \( x, y \in \mathbb{R} \). This is a well-known functional equation that suggests that \( f(x) \) could be an exponential function.

### Step 2: Finding the Form of \( f(x) \)
We also have the condition that \( f(x) = 1 + xg(x) \) where \( \lim_{x \to 0} g(x) = 1 \).

### Step 3: Evaluating \( f(0) \)
To find \( f(0) \), we can set \( x = 0 \) in the functional equation:
\[ f(0 + 0) = f(0)f(0) \implies f(0) = f(0)^2 \]
This implies \( f(0) = 0 \) or \( f(0) = 1 \). Since \( f(x) = 1 + xg(x) \) and \( \lim_{x \to 0} g(x) = 1 \), we have:
\[ f(0) = 1 + 0 \cdot g(0) = 1 \]
Thus, \( f(0) = 1 \).

### Step 4: Finding the Derivative \( f'(x) \)
We want to find \( f'(x) \). By the definition of the derivative:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Using the functional equation, we can express \( f(x+h) \):
\[ f(x+h) = f(x)f(h) \]
Thus, we have:
\[ f'(x) = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} = \lim_{h \to 0} \frac{f(x)(f(h) - 1)}{h} \]

### Step 5: Evaluating \( f(h) \)
From the expression for \( f(x) \):
\[ f(h) = 1 + hg(h) \]
Substituting this into our limit gives:
\[ f'(x) = \lim_{h \to 0} \frac{f(x)(1 + hg(h) - 1)}{h} = \lim_{h \to 0} \frac{f(x)hg(h)}{h} = f(x) \lim_{h \to 0} g(h) \]

### Step 6: Applying the Limit
Since we know that \( \lim_{h \to 0} g(h) = 1 \):
\[ f'(x) = f(x) \cdot 1 = f(x) \]

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

### Final Answer
The derivative \( f'(x) \) is equal to \( f(x) \).

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