A function `f: R->R` satisfies that equation `f(x+y)=f(x)f(y)` for all `x ,\ y in R` , `f(x)!=0` . Suppose that the function `f(x)` is differentiable at `x=0` and `f^(prime)(0)=2` . Prove that `f^(prime)(x)=2\ f(x)` .

Explanation

To solve the problem, we need to prove that if a function \( f: \mathbb{R} \to \mathbb{R} \) satisfies the equation

\[ f(x+y) = f(x)f(y) \]

for all \( x, y \in \mathbb{R} \) and \( f(x) \neq 0 \), and if \( f \) is differentiable at \( x = 0 \) with \( f'(0) = 2 \), then it follows that \( f'(x) = 2 f(x) \) for all \( x \).

### Step-by-Step Solution

1. **Understanding the Functional Equation**:
Given \( f(x+y) = f(x)f(y) \), we can set \( y = 0 \):
\[ f(x+0) = f(x)f(0) \implies f(x) = f(x)f(0) \]
Since \( f(x) \neq 0 \), we can divide both sides by \( f(x) \):
\[ 1 = f(0) \implies f(0) = 1 \]

2. **Finding \( f'(0) \)**:
We know that \( f'(0) = 2 \). The derivative at \( x = 0 \) is defined as:
\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h) - 1}{h} \]
Given that \( f'(0) = 2 \), we have:
\[ 2 = \lim_{h \to 0} \frac{f(h) - 1}{h} \]

3. **Finding \( f'(x) \)**:
To find \( f'(x) \) for any \( x \), we use the definition of the derivative:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Using the functional equation, we can express \( f(x+h) \):
\[ f(x+h) = f(x)f(h) \]
Thus,
\[ f'(x) = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} = \lim_{h \to 0} \frac{f(x)(f(h) - 1)}{h} \]
Since \( f(x) \) is constant with respect to \( h \), we can factor it out:
\[ f'(x) = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h} \]

4. **Using the Limit**:
From our earlier calculation, we know:
\[ \lim_{h \to 0} \frac{f(h) - 1}{h} = 2 \]
Therefore:
\[ f'(x) = f(x) \cdot 2 = 2f(x) \]

5. **Conclusion**:
We have shown that:
\[ f'(x) = 2f(x) \]
for all \( x \in \mathbb{R} \).

### Final Result
Thus, we conclude that \( f'(x) = 2f(x) \).