Let `f(x+y)=f(x)f(y)` and where `f(x)=1+xg(x)phi(x)`, `lim_(xto0)g(x)=a` and `lim_(xto0)phi(x)=b` Then what is `f'(x)` equal to ?

Options

A. `a+abf(x)`
B. `1+ab`
C. ab
D. `abf(x)`

Answer

To find \( f'(x) \) given the functional equation \( f(x+y) = f(x)f(y) \) and the form \( f(x) = 1 + xg(x)\phi(x) \), we can follow these steps:

### Step 1: Differentiate \( f(x) \)

We start with the definition of the derivative:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Use the functional equation

Using the property \( f(x+y) = f(x)f(y) \), we can express \( f(x+h) \):
\[ f(x+h) = f(x)f(h) \]

### Step 3: Substitute into the derivative formula

Substituting this into the derivative formula gives:
\[ f'(x) = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} \]

### Step 4: Factor out \( f(x) \)

Factoring out \( f(x) \) from the numerator:
\[ f'(x) = \lim_{h \to 0} f(x) \frac{f(h) - 1}{h} \]

### Step 5: Find \( f(h) - 1 \)

From the given form of \( f(x) \):
\[ f(h) = 1 + hg(h)\phi(h) \]
Thus:
\[ f(h) - 1 = hg(h)\phi(h) \]

### Step 6: Substitute back into the limit

Now substituting back, we have:
\[ f'(x) = \lim_{h \to 0} f(x) \frac{hg(h)\phi(h)}{h} \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} f(x) g(h) \phi(h) \]

### Step 7: Evaluate the limits

Using the limits given in the problem:
\[ \lim_{h \to 0} g(h) = a \quad \text{and} \quad \lim_{h \to 0} \phi(h) = b \]
We can substitute these limits into our expression:
\[ f'(x) = f(x) \cdot a \cdot b \]

### Step 8: Substitute \( f(x) \)

Now, substituting \( f(x) = 1 + xg(x)\phi(x) \) back into the expression:
\[ f'(x) = (1 + xg(x)\phi(x)) \cdot ab \]

### Final Result

Thus, we have:
\[ f'(x) = ab(1 + xg(x)\phi(x)) \]