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 need to analyze the given information and apply the properties of differentiable functions.

### Step 1: Understand the given function
We are given that \( f(xy) = f(x)f(y) \) for all \( x, y \in \mathbb{R} \). This is a functional equation that suggests that \( f \) could be an exponential function.

### Step 2: Differentiate the function
We know that \( f \) is differentiable at \( x = 1 \) and that \( f'(1) = 1 \). We also have \( f(1) \neq 0 \) and \( f(2) = 3 \).

### Step 3: Use the functional equation
Let's differentiate the functional equation with respect to \( x \) at \( x = 1 \) and \( y = 1 \):
\[ f'(xy) \cdot y = f'(x)f(y) + f(x)f'(y) \]
Substituting \( x = 1 \) and \( y = 1 \):
\[ f'(1 \cdot 1) \cdot 1 = f'(1)f(1) + f(1)f'(1) \]
This simplifies to:
\[ f'(1) = 2f'(1)f(1) \]
Since \( f'(1) = 1 \), we can substitute:
\[ 1 = 2 \cdot 1 \cdot f(1) \implies f(1) = \frac{1}{2} \]

### Step 4: Find \( f(2) \)
We know \( f(2) = 3 \). We can use the functional equation again:
\[ f(2) = f(1 \cdot 2) = f(1)f(2) \]
Substituting the known values:
\[ 3 = \frac{1}{2} \cdot f(2) \]
This gives:
\[ 3 = \frac{1}{2} \cdot 3 \implies \text{This is consistent.} \]

### Step 5: Find \( f'(2) \)
Now we need to find \( f'(2) \). We can use the limit definition of the derivative:
\[ f'(2) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} \]
Using the functional equation:
\[ f(2 + h) = f(2)f(1 + \frac{h}{2}) \text{ (by substituting \( x = 2 \) and \( y = 1 + \frac{h}{2} \))} \]
Thus:
\[ f'(2) = \lim_{h \to 0} \frac{f(2)f(1 + \frac{h}{2}) - f(2)}{h} \]
Factoring out \( f(2) \):
\[ = f(2) \lim_{h \to 0} \frac{f(1 + \frac{h}{2}) - 1}{h} \]
Using the substitution \( k = \frac{h}{2} \) (thus \( h = 2k \)):
\[ = f(2) \cdot 2 \cdot f'(1) \]
Substituting \( f(2) = 3 \) and \( f'(1) = 1 \):
\[ f'(2) = 3 \cdot 2 \cdot 1 = 6 \]

### Final Answer:
Thus, \( f'(2) = 6 \).