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 start with the functional equation given:

1. **Functional Equation**:
\[ f(x+y) = f(x)f(y) \quad \text{for all } x, y \in \mathbb{R} \]
We also know that \( f(0) \neq 0 \).

2. **Finding \( f(0) \)**:
By substituting \( x = 0 \) and \( y = 0 \) into the functional equation, we get:
\[ f(0 + 0) = f(0)f(0) \implies f(0) = f(0)^2 \]
This implies \( f(0)(1 - f(0)) = 0 \). Since \( f(0) \neq 0 \), we conclude that:
\[ f(0) = 1 \]

3. **Differentiating the Functional Equation**:
We differentiate both sides of the equation \( f(x+y) = f(x)f(y) \) with respect to \( y \):
\[ \frac{d}{dy} f(x+y) = f'(x+y) \]
Using the product rule on the right side:
\[ \frac{d}{dy} [f(x)f(y)] = f(x)f'(y) \]
Setting \( y = 0 \):
\[ f'(x) = f(x)f'(0) \]

4. **Using Given Information**:
We know \( f'(0) = 0 \). Thus:
\[ f'(x) = f(x) \cdot 0 = 0 \quad \text{for all } x \]
This means that \( f'(x) = 0 \) for all \( x \), indicating that \( f(x) \) is a constant function.

5. **Finding the Constant Value**:
Since \( f(5) = 10 \), and \( f(x) \) is constant, we conclude:
\[ f(x) = 10 \quad \text{for all } x \]

6. **Finding \( f'(5) \)**:
Since \( f(x) \) is constant, its derivative is:
\[ f'(5) = 0 \]

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