If `f : R-{-1}->R ` and `f` is differentiable function satisfies `f(x+f(y) +x f(y)) = y+f(x)+y.f(x) forall x,y in R-{-1}` then find the value of `2010 [ 1 +f(2009)]`

Answer

To solve the problem, we need to analyze the given functional equation step by step. The equation we are given is:

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

for all \( x, y \in \mathbb{R} \setminus \{-1\} \).

### Step 1: Substitute \( y = 0 \)

Let's first substitute \( y = 0 \) into the equation:

\[ f(x + f(0) + x f(0)) = 0 + f(x) + 0 \cdot f(x) \]

This simplifies to:

\[ f(x + f(0) + x f(0)) = f(x) \]

### Step 2: Set \( f(0) = c \)

Let \( f(0) = c \). Then we have:

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

This implies that \( f \) is periodic with respect to the argument \( x + c + x c \).

### Step 3: Substitute \( x = 0 \)

Now, substitute \( x = 0 \) into the original equation:

\[ f(0 + f(y) + 0 \cdot f(y)) = y + f(0) + y \cdot f(0) \]

This simplifies to:

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

### Step 4: Analyze \( f(f(y)) \)

From the previous step, we can express \( f(f(y)) \):

\[ f(f(y)) = y(1 + c) + c \]

### Step 5: Set \( c = 0 \)

Assuming \( c = 0 \) (since \( f(0) = 0 \)), we have:

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

This indicates that \( f \) is an involution, meaning \( f(f(y)) = y \).

### Step 6: Substitute \( y = x \)

Now, substitute \( y = x \) into the original equation:

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

This means:

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

### Step 7: Conclude that \( f(x) = x \)

Given that \( f(f(y)) = y \) and \( f(0) = 0 \), we can conclude that \( f(x) = x \) for all \( x \).

### Step 8: Calculate \( 2010(1 + f(2009)) \)

Now, we need to find the value of \( 2010(1 + f(2009)) \):

\[ f(2009) = 2009 \]

Thus:

\[ 2010(1 + f(2009)) = 2010(1 + 2009) = 2010 \cdot 2010 = 4040100 \]

### Final Answer

The final value is:

\[ \boxed{4040100} \]