Let `f : R^(+) rarr R` satisfies the functional equation `f(xy) = e^(xy - x - y) {e^(y) f(x) + e^(x) f(y)}, AA x, y in R^(+)`. If f'(1) = e, determine f(x).

Answer

To solve the problem, we will analyze the functional equation step by step and find the function \( f(x) \).

### Step 1: Substitute \( y = 1 \)

We start by substituting \( y = 1 \) into the functional equation:

\[ f(xy) = e^{xy - x - y} \left( e^y f(x) + e^x f(y) \right) \]

This gives us:

\[ f(x \cdot 1) = e^{x \cdot 1 - x - 1} \left( e^1 f(x) + e^x f(1) \right) \]

Which simplifies to:

\[ f(x) = e^{x - x - 1} \left( e f(x) + e^x f(1) \right) \]

This further simplifies to:

\[ f(x) = e^{-1} \left( e f(x) + e^x f(1) \right) \]

### Step 2: Rearranging the equation

Now, we can rearrange the equation:

\[ f(x) = \frac{1}{e} \left( e f(x) + e^x f(1) \right) \]

Multiplying both sides by \( e \):

\[ e f(x) = e f(x) + e^x f(1) \]

Subtracting \( e f(x) \) from both sides:

\[ 0 = e^x f(1) \]

### Step 3: Analyzing the result

Since \( e^x \) is never zero for any \( x \in \mathbb{R}^+ \), we conclude that:

\[ f(1) = 0 \]

### Step 4: Substitute \( f(1) \) back into the equation

Now we will substitute \( f(1) = 0 \) back into our original functional equation. We rewrite the equation as:

\[ f(xy) = e^{xy - x - y} \left( e^y f(x) + e^x f(y) \right) \]

### Step 5: Assume a form for \( f(x) \)

Let’s assume \( f(x) = e^x g(x) \) for some function \( g(x) \). Then we have:

\[ f(xy) = e^{xy} g(xy) \]

Substituting this into the functional equation gives:

\[ e^{xy} g(xy) = e^{xy - x - y} \left( e^y e^x g(x) + e^x e^y g(y) \right) \]

This simplifies to:

\[ g(xy) = g(x) + g(y) \]

This is Cauchy’s functional equation, which has the general solution of the form \( g(x) = k \log x \) for some constant \( k \).

### Step 6: Determine \( k \)

Thus, we have:

\[ f(x) = e^x k \log x \]

To find \( k \), we use the condition \( f'(1) = e \).

### Step 7: Differentiate \( f(x) \)

Differentiating \( f(x) \):

\[ f'(x) = e^x k \log x + e^x k \cdot \frac{1}{x} \]

Evaluating at \( x = 1 \):

\[ f'(1) = e^1 k \cdot 0 + e^1 k \cdot 1 = k e \]

Setting this equal to \( e \):

\[ k e = e \implies k = 1 \]

### Final Result

Thus, we find:

\[ f(x) = e^x \log x \]

### Summary

The function \( f(x) \) is:

\[ \boxed{e^x \log x} \]