Let `f(x)` is a differentiable function on `x in R`, such that `f(x+y)=f(x)f(y)` for all `x, y in R` where `f(0) ne 0`. If `f(5)=10, f'(0)=0`, then the value of `f'(5)` is equal to

Answer

To solve the problem, we need to find the value of \( f'(5) \) given the properties of the function \( f(x) \).

### Step-by-Step Solution

1. **Understanding the Functional Equation**:
We are given that \( f(x+y) = f(x)f(y) \) for all \( x, y \in \mathbb{R} \) and \( f(0) \neq 0 \). This is a well-known functional equation that suggests that \( f(x) \) could be an exponential function.

2. **Finding \( f(0) \)**:
Set \( x = 0 \) and \( y = 0 \):
\[ f(0 + 0) = f(0)f(0) \implies f(0) = f(0)^2 \]
Since \( f(0) \neq 0 \), we can divide both sides by \( f(0) \):
\[ 1 = f(0) \]
Thus, \( f(0) = 1 \).

3. **Differentiating the Functional Equation**:
Differentiate both sides of the equation \( f(x+y) = f(x)f(y) \) with respect to \( y \):
\[ \frac{d}{dy}f(x+y) = f'(x+y) \quad \text{and} \quad \frac{d}{dy}(f(x)f(y)) = f(x)f'(y) \]
Setting \( y = 0 \):
\[ f'(x) = f(x)f'(0) \]

4. **Using Given Information**:
We know \( f'(0) = 0 \). Thus:
\[ f'(x) = f(x) \cdot 0 = 0 \]
This means that \( f'(x) = 0 \) for all \( x \).

5. **Finding \( f'(5) \)**:
Since \( f'(x) = 0 \) for all \( x \), we specifically have:
\[ f'(5) = 0 \]

### Conclusion
Therefore, the value of \( f'(5) \) is \( \boxed{0} \).