Let f(x + y)=f(x)+f (y) and `f(x) = x^2 g(x)` for all `x, y in R`, where g(x) is continuous function. Then f'(x) is equal to

Options

A. g'(x)
B. g(0)
C. g(0) + g'(0)
D. 0

Answer

To solve the problem, we need to find \( f'(x) \) given the conditions \( f(x + y) = f(x) + f(y) \) and \( f(x) = x^2 g(x) \), where \( g(x) \) is a continuous function.

### Step-by-Step Solution:

1. **Understanding the Functional Equation**:
The equation \( f(x + y) = f(x) + f(y) \) suggests that \( f \) is a Cauchy additive function. A common solution to this type of equation is of the form \( f(x) = kx \) for some constant \( k \), but here we have an additional condition on \( f(x) \).

2. **Substituting the Given Form of f**:
We know that \( f(x) = x^2 g(x) \). We can substitute this into the functional equation:
\[ f(x + y) = (x + y)^2 g(x + y) = f(x) + f(y) = x^2 g(x) + y^2 g(y) \]

3. **Expanding the Left Side**:
Expanding the left side gives:
\[ (x + y)^2 g(x + y) = (x^2 + 2xy + y^2) g(x + y) \]

4. **Equating Both Sides**:
Thus, we have:
\[ (x^2 + 2xy + y^2) g(x + y) = x^2 g(x) + y^2 g(y) \]

5. **Analyzing the Equation**:
For this equation to hold for all \( x \) and \( y \), we can set \( y = 0 \):
\[ (x^2 + 0) g(x) = x^2 g(x) + 0 \]
This is trivially true. Now, let's differentiate \( f(x) \).

6. **Finding the Derivative**:
We can find \( f'(x) \) using the product rule:
\[ f'(x) = \frac{d}{dx}(x^2 g(x)) = 2x g(x) + x^2 g'(x) \]

7. **Using the Limit Definition**:
We can also find \( f'(x) \) using the limit definition:
\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]
Substituting \( f(x + h) = (x + h)^2 g(x + h) \) and \( f(x) = x^2 g(x) \):
\[ f'(x) = \lim_{h \to 0} \frac{(x + h)^2 g(x + h) - x^2 g(x)}{h} \]

8. **Simplifying the Expression**:
Expanding \( (x + h)^2 \):
\[ f'(x) = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2) g(x + h) - x^2 g(x)}{h} \]
This can be split into two parts:
\[ = \lim_{h \to 0} \left( \frac{2xh g(x + h)}{h} + \frac{h^2 g(x + h)}{h} + \frac{x^2 (g(x + h) - g(x))}{h} \right) \]

9. **Taking the Limit**:
As \( h \to 0 \), \( g(x + h) \to g(x) \):
\[ f'(x) = 2x g(x) + 0 + x^2 g'(x) = 2x g(x) + x^2 g'(x) \]

### Final Answer:
Thus, the derivative \( f'(x) \) is given by:
\[ f'(x) = 2x g(x) + x^2 g'(x) \]