Let `f(x)=x^n` ,`n` being a non negative integer. The value of `n` for which the equality`f'(a+b)=f'(a)+f'(b)` is valid for all `a.bgt0` is

Answer

To solve the problem, we need to find the value of \( n \) such that the equality \( f'(a+b) = f'(a) + f'(b) \) holds for the function \( f(x) = x^n \) where \( n \) is a non-negative integer and \( a, b > 0 \).

### 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. **Evaluate the derivatives at \( a+b \), \( a \), and \( b \)**:
- For \( f'(a+b) \):
\[ f'(a+b) = n \cdot (a+b)^{n-1} \]
- For \( f'(a) \):
\[ f'(a) = n \cdot a^{n-1} \]
- For \( f'(b) \):
\[ f'(b) = n \cdot b^{n-1} \]

3. **Set up the equation**:
We need to satisfy the equation:
\[ f'(a+b) = f'(a) + f'(b) \]
Substituting the derivatives we found:
\[ n \cdot (a+b)^{n-1} = n \cdot a^{n-1} + n \cdot b^{n-1} \]

4. **Simplify the equation**:
We can divide both sides by \( n \) (assuming \( n \neq 0 \)):
\[ (a+b)^{n-1} = a^{n-1} + b^{n-1} \]

5. **Analyze the equation**:
The equation \( (a+b)^{n-1} = a^{n-1} + b^{n-1} \) must hold for all \( a, b > 0 \). This equality is known to hold true for \( n-1 = 1 \), which implies:
\[ n - 1 = 1 \implies n = 2 \]

6. **Conclusion**:
Thus, the value of \( n \) for which the equality \( f'(a+b) = f'(a) + f'(b) \) is valid for all \( a, b > 0 \) is:
\[ \boxed{2} \]