Let `f(x+y)=f(x).f(y)` for all `x, y in R` and `f(x)=1+x phi(x)log3.` If `lim_(xrarr0)phi(x)=1,` then f'(x) is equal to

Options

A. `log3^(f(x))`
B. `log[f(x)]^(3)`
C. `log3`
D. None of these

Answer

To solve the problem, we start with the given functional equation and the expression for \( f(x) \).

1. **Given Functional Equation**:
\[ f(x+y) = f(x) \cdot f(y) \quad \text{for all } x, y \in \mathbb{R} \]
This indicates that \( f(x) \) is an exponential function.

2. **Given Function**:
\[ f(x) = 1 + x \phi(x) \log 3 \]
where \( \lim_{x \to 0} \phi(x) = 1 \).

3. **Finding \( f'(x) \)**:
We will differentiate \( f(x) \) using the definition of the derivative:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

4. **Substituting in the Functional Equation**:
Using the functional equation, we have:
\[ f(x+h) = f(x) \cdot f(h) \]
Thus,
\[ f'(x) = \lim_{h \to 0} \frac{f(x) \cdot f(h) - f(x)}{h} \]
This simplifies to:
\[ f'(x) = f(x) \cdot \lim_{h \to 0} \frac{f(h) - 1}{h} \]

5. **Finding \( f(h) \)**:
From the given function:
\[ f(h) = 1 + h \phi(h) \log 3 \]
Therefore,
\[ f(h) - 1 = h \phi(h) \log 3 \]

6. **Substituting Back**:
Now substituting this back into the limit:
\[ \lim_{h \to 0} \frac{f(h) - 1}{h} = \lim_{h \to 0} \phi(h) \log 3 \]
Given that \( \lim_{h \to 0} \phi(h) = 1 \), we have:
\[ \lim_{h \to 0} \phi(h) \log 3 = \log 3 \]

7. **Final Expression for \( f'(x) \)**:
Thus, we can write:
\[ f'(x) = f(x) \cdot \log 3 \]

8. **Substituting \( f(x) \)**:
Now substituting \( f(x) = 1 + x \phi(x) \log 3 \):
\[ f'(x) = (1 + x \phi(x) \log 3) \cdot \log 3 \]

9. **Conclusion**:
Therefore, the final expression for \( f'(x) \) is:
\[ f'(x) = (1 + x \phi(x) \log 3) \cdot \log 3 \]