Suppose the function f satisfies the conditions :
(i) `f(x+y) =f(x) f(y)` for all x and y.
(ii) `f(x) = 1 + xg(x)`, where `lim_(x to 0) g(x)=1`
Then `f'(x)` is:

Options

A. `f(x)`
B. `f(x) g(0)`
C. `2f(x) g(0)`
D. none

Answer

To find \( f'(x) \) given the conditions of the function \( f \), we can proceed as follows:

### Step 1: Understand the properties of the function
We know that:
1. \( f(x+y) = f(x) f(y) \) for all \( x \) and \( y \).
2. \( f(x) = 1 + x g(x) \) where \( \lim_{x \to 0} g(x) = 1 \).

### Step 2: Differentiate \( f(x) \)
To find \( f'(x) \), we will use the definition of the derivative:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 3: Apply the first condition
Using the first condition \( f(x+h) = f(x) f(h) \):
\[ f'(x) = \lim_{h \to 0} \frac{f(x) f(h) - f(x)}{h} \]
This can be factored as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x)(f(h) - 1)}{h} \]

### Step 4: Simplify using the second condition
Now we substitute \( f(h) \) using the second condition:
\[ f(h) = 1 + h g(h) \]
Thus,
\[ f(h) - 1 = h g(h) \]
Substituting this back into our expression for \( f'(x) \):
\[ f'(x) = \lim_{h \to 0} \frac{f(x) (h g(h))}{h} \]
The \( h \) in the numerator and denominator cancels out:
\[ f'(x) = \lim_{h \to 0} f(x) g(h) \]

### Step 5: Evaluate the limit
Since \( \lim_{h \to 0} g(h) = 1 \):
\[ f'(x) = f(x) \cdot 1 = f(x) \]

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

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

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