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. 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 \( f(x) = x^n \) and \( n \) is a non-negative integer.

### Step-by-Step Solution:

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

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

3. **Calculate \( f'(x) + f'(y) \):**
Now, we calculate \( f'(x) + f'(y) \):
\[ f'(x) + f'(y) = n x^{n-1} + n y^{n-1} = n (x^{n-1} + y^{n-1}) \]

4. **Set up the equation:**
We need to find \( n \) such that:
\[ n (x+y)^{n-1} = n (x^{n-1} + y^{n-1}) \]
Assuming \( n \neq 0 \), we can divide both sides by \( n \):
\[ (x+y)^{n-1} = x^{n-1} + y^{n-1} \]

5. **Analyze the equation:**
We need to check for which values of \( n \) this equation holds for all \( x, y > 0 \).

6. **Test specific values of \( n \):**
- For \( n = 0 \):
\[ (x+y)^{-1} \quad \text{(LHS)} \quad \text{and} \quad x^{-1} + y^{-1} \quad \text{(RHS)} \]
This gives:
\[ \frac{1}{x+y} \neq \frac{1}{x} + \frac{1}{y} \]
So, \( n = 0 \) is not a solution.

- For \( n = 1 \):
\[ (x+y)^{0} = 1 \quad \text{(LHS)} \quad \text{and} \quad x^{0} + y^{0} = 1 + 1 = 2 \quad \text{(RHS)} \]
So, \( n = 1 \) is not a solution.

- For \( n = 2 \):
\[ (x+y)^{1} = x+y \quad \text{(LHS)} \quad \text{and} \quad x^{1} + y^{1} = x + y \quad \text{(RHS)} \]
This holds true, so \( n = 2 \) is a solution.

- For \( n \geq 3 \):
The left-hand side \( (x+y)^{n-1} \) grows faster than the right-hand side \( x^{n-1} + y^{n-1} \) for \( x, y > 0 \), hence they will not be equal.

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