Let `f:R to R` be given by `f(x+y)=f(x)-f(y)+2xy+1"for all "x,y in R` If f(x) is everywhere differentiable and `f'(0)=1`, then f'(x)=

Options

A. 2x+1
B. 2x-1
C. x+1
D. x-1

Explanation

To solve the problem, we start with the functional equation given:

\[ f(x+y) = f(x) - f(y) + 2xy + 1 \]

for all \( x, y \in \mathbb{R} \). We also know that \( f \) is everywhere differentiable and \( f'(0) = 1 \). We need to find \( f'(x) \).

### Step 1: Find \( f(0) \)

Let’s substitute \( x = 0 \) and \( y = 0 \) into the functional equation:

\[ f(0 + 0) = f(0) - f(0) + 2 \cdot 0 \cdot 0 + 1 \]

This simplifies to:

\[ f(0) = 0 + 1 \implies f(0) = 1 \]

### Step 2: Differentiate the functional equation

Next, we differentiate both sides of the functional equation with respect to \( y \):

\[ \frac{d}{dy} f(x+y) = \frac{d}{dy} \left( f(x) - f(y) + 2xy + 1 \right) \]

Using the chain rule on the left side, we have:

\[ f'(x+y) = -f'(y) + 2x \]

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

Now, let’s substitute \( y = 0 \):

\[ f'(x+0) = -f'(0) + 2x \]

This simplifies to:

\[ f'(x) = -f'(0) + 2x \]

Given that \( f'(0) = 1 \), we substitute this value in:

\[ f'(x) = -1 + 2x \]

### Step 4: Final expression for \( f'(x) \)

Thus, we can write:

\[ f'(x) = 2x - 1 \]

### Conclusion

The derivative of the function \( f \) is:

\[ f'(x) = 2x - 1 \]