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 prove that \( f'(x) = 2f(x) \) given the functional equation \( f(x+y) = f(x)f(y) \) and the conditions \( f(0) \neq 0 \) and \( f'(0) = 2 \), we can follow these steps:

### Step 1: Analyze the Functional Equation
We start with the functional equation:
\[ f(x+y) = f(x)f(y) \]
Setting \( y = 0 \), we have:
...