Let `f(x y)=f(x)f(y)AAx , y in Ra n df` is differentiable at `x=1` such that `f^(prime)(1)=1.` Also, `f(1)!=0,f(2)=3.` Then find `f^(prime)(2)dot`

Explanation

To solve the problem step by step, we will use the properties of the function \( f(x) \) and the given information.

### Step-by-Step Solution:

1. **Given Functional Equation**: We have the equation \( f(xy) = f(x)f(y) \) for all \( x, y \in \mathbb{R} \).

2. **Differentiate the Functional Equation**: We will differentiate both sides of the equation with respect to \( y \) while treating \( x \) as a constant.
\[ \frac{d}{dy}[f(xy)] = \frac{d}{dy}[f(x)f(y)] \]
Using the chain rule on the left side, we get:
\[ f'(xy) \cdot x = f(x)f'(y) \]

3. **Substituting Values**: Now, let’s substitute \( x = 1 \) and \( y = 1 \):
\[ f'(1 \cdot 1) \cdot 1 = f(1)f'(1) \]
This simplifies to:
\[ f'(1) = f(1)f'(1) \]

4. **Using Given Information**: We know \( f'(1) = 1 \) and \( f(1) \neq 0 \). Thus, we can rearrange the equation:
\[ 1 = f(1) \cdot 1 \implies f(1) = 1 \]

5. **Finding \( f'(2) \)**: We need to find \( f'(2) \). We can express \( f'(2) \) using the definition of the derivative:
\[ f'(2) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} \]

6. **Using the Functional Equation Again**: We can rewrite \( f(2 + h) \) as \( f(2 \cdot (1 + \frac{h}{2})) \):
\[ f(2 + h) = f(2)f(1 + \frac{h}{2}) \]
Thus,
\[ f'(2) = \lim_{h \to 0} \frac{f(2)f(1 + \frac{h}{2}) - f(2)}{h} \]

7. **Factoring Out \( f(2) \)**: We can factor out \( f(2) \):
\[ f'(2) = f(2) \cdot \lim_{h \to 0} \frac{f(1 + \frac{h}{2}) - 1}{h} \]
The limit simplifies to:
\[ \lim_{k \to 0} \frac{f(1 + k) - 1}{2k} = \frac{1}{2} f'(1) = \frac{1}{2} \cdot 1 = \frac{1}{2} \]

8. **Final Calculation**: Now substituting \( f(2) = 3 \):
\[ f'(2) = 3 \cdot \frac{1}{2} = \frac{3}{2} \]

### Final Answer:
\[ f'(2) = \frac{3}{2} \]