If `f(x)=tan^(-1)x+cos^(-1)((1-x^(2))/(1+x^(2)))`, then

Options

A. `f'(-2)=(4)/(5)`
B. `f'(-1)=-1`
C. `f'(x)=0" for all "xlt0`
D. none of these

Explanation

To solve the problem, we need to analyze the function given:

\[ f(x) = \tan^{-1}(x) + \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \]

### Step 1: Simplifying the Function
We can simplify the term \(\cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right)\).

Using the identity:
\[ \cos^{-1}(y) = 2\tan^{-1}\left(\frac{\sqrt{1-y}}{\sqrt{1+y}}\right) \]
we can rewrite:
\[ \frac{1 - x^2}{1 + x^2} = \cos(2\theta) \quad \text{where } \tan(\theta) = x \]
This gives:
\[ \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) = 2\tan^{-1}(x) \]

Thus, we can rewrite \(f(x)\):
\[ f(x) = \tan^{-1}(x) + 2\tan^{-1}(x) = 3\tan^{-1}(x) \quad \text{for } x \geq 0 \]

For \(x < 0\), we have:
\[ \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) = -2\tan^{-1}(x) \]
So,
\[ f(x) = \tan^{-1}(x) - 2\tan^{-1}(x) = -\tan^{-1}(x) \quad \text{for } x < 0 \]

### Step 2: Finding the Derivative
Now, we differentiate \(f(x)\):
- For \(x \geq 0\):
\[ f'(x) = 3 \cdot \frac{1}{1 + x^2} \]
- For \(x < 0\):
\[ f'(x) = -\frac{1}{1 + x^2} \]

### Step 3: Evaluating the Derivative at Specific Points
1. **At \(x = -1\)**:
- Since \(-1 < 0\):
\[ f'(-1) = -\frac{1}{1 + (-1)^2} = -\frac{1}{2} \]

2. **At \(x = -2\)**:
- Since \(-2 < 0\):
\[ f'(-2) = -\frac{1}{1 + (-2)^2} = -\frac{1}{5} \]

3. **At \(x = 0\)**:
- Since \(0 \geq 0\):
\[ f'(0) = 3 \cdot \frac{1}{1 + 0^2} = 3 \]

### Conclusion
- For \(x = -1\), \(f'(-1) = -\frac{1}{2}\)
- For \(x = -2\), \(f'(-2) = -\frac{1}{5}\)
- For \(x = 0\), \(f'(0) = 3\)

### Final Answer
None of the options provided in the question are correct based on our calculations.