If `f(x)` is a real valued and differentaible function on R and `f(x+y)=(f(x)+f(y))/(1-f(x)f(y))`, then show that `f'(x)=f'(0)[1+t^(2)(x)].`

Answer

To solve the problem, we need to show that if \( f(x+y) = \frac{f(x) + f(y)}{1 - f(x)f(y)} \), then \( f'(x) = f'(0)(1 + f^2(x)) \).

### Step 1: Differentiate the given functional equation
We start with the equation:
\[ f(x+y) = \frac{f(x) + f(y)}{1 - f(x)f(y)} \]
We will differentiate both sides with respect to \( y \) at \( y = 0 \).

### Step 2: Apply the chain rule on the left-hand side
Using the chain rule, we differentiate the left-hand side:
\[ \frac{d}{dy} f(x+y) \bigg|_{y=0} = f'(x) \]

### Step 3: Differentiate the right-hand side
Now, we differentiate the right-hand side. Let \( g(y) = f(y) \). Then we have:
\[ \frac{d}{dy} \left( \frac{f(x) + g(y)}{1 - f(x)g(y)} \right) \bigg|_{y=0} \]
Using the quotient rule:
\[ \frac{u}{v} \Rightarrow \frac{u'v - uv'}{v^2} \]
where \( u = f(x) + g(y) \) and \( v = 1 - f(x)g(y) \).

### Step 4: Compute \( u' \) and \( v' \)
- \( u' = g'(y) = f'(y) \) and at \( y=0 \), \( g'(0) = f'(0) \).
- \( v' = -f(x)g'(y) \) and at \( y=0 \), \( v' = -f(x)f'(0) \).

### Step 5: Substitute \( y = 0 \) into the derivatives
Now substituting \( y = 0 \):
- \( u = f(x) + f(0) \)
- \( v = 1 - f(x)f(0) \)

Thus, we have:
\[ \frac{d}{dy} \left( \frac{f(x) + f(y)}{1 - f(x)f(y)} \right) \bigg|_{y=0} = \frac{f'(0)(1 - f(x)f(0)) - (f(x) + f(0))(-f(x)f'(0))}{(1 - f(x)f(0))^2} \]

### Step 6: Simplify the expression
This simplifies to:
\[ f'(x) = \frac{f'(0)(1 - f(x)f(0)) + f'(0)f(x)(f(x) + f(0))}{(1 - f(x)f(0))^2} \]

### Step 7: Analyze the limit as \( f(0) \to 0 \)
Assuming \( f(0) = 0 \), we can simplify further:
\[ f'(x) = f'(0)(1 + f^2(x)) \]

### Conclusion
Thus, we have shown that:
\[ f'(x) = f'(0)(1 + f^2(x)) \]