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)+x^(2) g(x) phi(x)"such that "lim_(x to 0) g(x)=a and lim_(x to 0) phi(x)=b,` then f'(x) is equal to

Options

A. `(a+b)f(x)`
B. af(x)
C. bf (x)
D. abf (x)

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Functional Equation**:
The functional equation \( f(x+y) = f(x)f(y) \) suggests that \( f(x) \) is an exponential function. A common solution to this type of equation is \( f(x) = e^{kx} \) for some constant \( k \). However, we will work with the provided form of \( f(x) \).

2. **Given Form of the Function**:
We have:
\[ f(x) = 1 + xg(x) + x^2g(x)\phi(x) \]
where \( \lim_{x \to 0} g(x) = a \) and \( \lim_{x \to 0} \phi(x) = b \).

3. **Finding \( f'(x) \)**:
To find \( f'(x) \), we will differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(1 + xg(x) + x^2g(x)\phi(x)) \]
Using the product rule and chain rule:
\[ f'(x) = g(x) + xg'(x) + \left(2xg(x)\phi(x) + x^2g(x)\phi'(x)\right) \]

4. **Evaluating the Limit as \( x \to 0 \)**:
We will evaluate \( f'(x) \) as \( x \to 0 \):
- As \( x \to 0 \), \( g(x) \to a \), \( g'(x) \to g'(0) \), \( \phi(x) \to b \), and \( \phi'(x) \to \phi'(0) \).
- Thus, we have:
\[ f'(0) = g(0) + 0 \cdot g'(0) + \left(2 \cdot 0 \cdot a \cdot b + 0^2 \cdot a \cdot \phi'(0)\right) = a \]

5. **Using the Functional Equation**:
From the functional equation, we know:
\[ f'(x) = f(x) \cdot \lim_{h \to 0} \frac{f(h) - 1}{h} \]
Since \( f(h) \) approaches \( 1 \) as \( h \to 0 \), we find:
\[ \lim_{h \to 0} \frac{f(h) - 1}{h} = g(0) = a \]
Therefore:
\[ f'(x) = f(x) \cdot a \]

6. **Final Expression for \( f'(x) \)**:
Thus, we conclude:
\[ f'(x) = a f(x) \]

### Conclusion:
The derivative \( f'(x) \) is equal to \( a f(x) \).