Let `f(x)=x^(n)` , n being a non-negative integer, The value of `n` for which the equality `f'(x+y)=f'(x)+f'(y)` is valid for all `x,ygt0,` is

Options

A. 0,1
B. 1,2
C. 2,4
D. None of these

Answer

To solve the problem, we need to find the value of \( n \) for which the equality \( f'(x+y) = f'(x) + f'(y) \) holds for all \( x, y > 0 \), where \( f(x) = x^n \).

### Step-by-Step Solution:

1. **Define the function and its derivative**:
\[ f(x) = x^n \]
The derivative of \( f(x) \) is given by:
\[ f'(x) = n \cdot x^{n-1} \]

2. **Calculate \( f'(x+y) \)**:
Using the definition of the derivative, we have:
\[ f'(x+y) = n \cdot (x+y)^{n-1} \]

3. **Calculate \( f'(x) + f'(y) \)**:
Using the derivative calculated earlier:
\[ f'(x) + f'(y) = n \cdot x^{n-1} + n \cdot y^{n-1} = n \cdot (x^{n-1} + y^{n-1}) \]

4. **Set up the equation**:
We need to equate \( f'(x+y) \) and \( f'(x) + f'(y) \):
\[ n \cdot (x+y)^{n-1} = n \cdot (x^{n-1} + y^{n-1}) \]

5. **Assuming \( n \neq 0 \)** (we can divide by \( n \)):
\[ (x+y)^{n-1} = x^{n-1} + y^{n-1} \]

6. **Check for specific values of \( n \)**:
- **For \( n = 1 \)**:
\[ (x+y)^{1-1} = x^{1-1} + y^{1-1} \implies 1 = 1 + 1 \quad \text{(False)} \]
- **For \( n = 2 \)**:
\[ (x+y)^{2-1} = x^{2-1} + y^{2-1} \implies (x+y) = x + y \quad \text{(True)} \]
- **For \( n = 3 \)**:
\[ (x+y)^{3-1} = x^{3-1} + y^{3-1} \implies (x+y)^2 = x^2 + y^2 \quad \text{(False)} \]
- **For \( n = 4 \)**:
\[ (x+y)^{4-1} = x^{4-1} + y^{4-1} \implies (x+y)^3 = x^3 + y^3 \quad \text{(False)} \]

7. **Conclusion**:
The only value of \( n \) for which the equality \( f'(x+y) = f'(x) + f'(y) \) holds for all \( x, y > 0 \) is:
\[ n = 2 \]