Let f(x) be the function given by `f(x)=arc cos ((1-x^(2))/(1+x^(2))).Then`

Options

A. f(x) is everywhere differential such that `f'(x)=(2)/(1+x^(2))`
B. `f'(x)={{:(,(2)/(1+x^(2)),x gt0),(,(-2)/(1+x^(2)),x lt 0):}`
C. `f'(x)={{:(,(-2)/(1+x^(2)),x gt0),(,(2)/(1+x^(2)),x lt 0):}`
D. f'(x) exists at x=0

Explanation

To solve the problem, we need to differentiate the given function \( f(x) = \arccos\left(\frac{1 - x^2}{1 + x^2}\right) \). Let's go through the solution step by step.

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \arccos\left(\frac{1 - x^2}{1 + x^2}\right) \]

### Step 2: Differentiate using the chain rule
To differentiate \( f(x) \), we apply the chain rule. The derivative of \( \arccos(u) \) is given by:
\[ \frac{d}{dx} \arccos(u) = -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
where \( u = \frac{1 - x^2}{1 + x^2} \).

### Step 3: Find \( \frac{du}{dx} \)
Now, we need to differentiate \( u \):
\[ u = \frac{1 - x^2}{1 + x^2} \]
Using the quotient rule:
\[ \frac{du}{dx} = \frac{(1 + x^2)(-2x) - (1 - x^2)(2x)}{(1 + x^2)^2} \]
Simplifying this:
\[ \frac{du}{dx} = \frac{-2x(1 + x^2) - 2x(1 - x^2)}{(1 + x^2)^2} = \frac{-2x(1 + x^2 + 1 - x^2)}{(1 + x^2)^2} = \frac{-4x}{(1 + x^2)^2} \]

### Step 4: Substitute \( u \) and \( \frac{du}{dx} \) into the derivative
Now we substitute \( u \) and \( \frac{du}{dx} \) back into the derivative of \( f(x) \):
\[ f'(x) = -\frac{1}{\sqrt{1 - \left(\frac{1 - x^2}{1 + x^2}\right)^2}} \cdot \frac{-4x}{(1 + x^2)^2} \]

### Step 5: Simplify \( 1 - u^2 \)
Next, we need to simplify \( 1 - u^2 \):
\[ 1 - u^2 = 1 - \left(\frac{1 - x^2}{1 + x^2}\right)^2 = \frac{(1 + x^2)^2 - (1 - x^2)^2}{(1 + x^2)^2} \]
Calculating the numerator:
\[ (1 + x^2)^2 - (1 - x^2)^2 = (1 + 2x^2 + x^4) - (1 - 2x^2 + x^4) = 4x^2 \]
Thus:
\[ 1 - u^2 = \frac{4x^2}{(1 + x^2)^2} \]

### Step 6: Substitute back into \( f'(x) \)
Now substituting back into the derivative:
\[ f'(x) = -\frac{1}{\sqrt{\frac{4x^2}{(1 + x^2)^2}}} \cdot \frac{-4x}{(1 + x^2)^2} = \frac{(1 + x^2)}{2|x|} \cdot \frac{4x}{(1 + x^2)^2} \]
This simplifies to:
\[ f'(x) = \frac{2}{1 + x^2} \quad \text{for } x > 0 \]
And for \( x < 0 \), we have:
\[ f'(x) = -\frac{2}{1 + x^2} \]

### Step 7: Conclusion
Thus, we conclude:
- For \( x > 0 \), \( f'(x) = \frac{2}{1 + x^2} \)
- For \( x < 0 \), \( f'(x) = -\frac{2}{1 + x^2} \)

### Final Answer
The correct option is:
- \( f'(x) = \frac{2}{1 + x^2} \) for \( x > 0 \)
- \( f'(x) = -\frac{2}{1 + x^2} \) for \( x < 0 \)