If `f(x)=tan^(-1) ((2x)/(1-x^(2))), x in R "then "f'(x)` is given by

Options

A. `f'(x)=(2)/(1+x^(2))"for all x"in R (-1,1)`
B. `f'(x)=(2)/(1+x^(2))"for all x"in R`
C. `F'(x)={{:(,(2)/(1+x^(2)),"if "|x| le 1),(,(-2)/(1+x^(2)),"if |x| gt 1):}`
D. `f'(x)={{:(,(2)/(1+x^(2)),"if "|x| lt 1),(,(-2)/(1+x^(2)),"if |x| gt 1):}`

Explanation

To find the derivative \( f'(x) \) of the function \( f(x) = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \), we will follow these steps:

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

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

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

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

### Step 5: Simplify \( 1 + u^2 \)
Calculating \( 1 + u^2 \):
\[ 1 + u^2 = 1 + \left( \frac{2x}{1 - x^2} \right)^2 = 1 + \frac{4x^2}{(1 - x^2)^2} = \frac{(1 - x^2)^2 + 4x^2}{(1 - x^2)^2} = \frac{1 - 2x^2 + x^4 + 4x^2}{(1 - x^2)^2} = \frac{1 + 2x^2 + x^4}{(1 - x^2)^2} \]

### Step 6: Final expression for \( f'(x) \)
Now substituting back, we get:
\[ f'(x) = \frac{2(1 + x^2)}{(1 + 2x^2 + x^4)} \cdot \frac{(1 - x^2)^2}{(1 - x^2)^2} \]
This simplifies to:
\[ f'(x) = \frac{2(1 + x^2)}{1 + 2x^2 + x^4} \]

### Step 7: Check points of non-differentiability
The function \( f(x) \) is not continuous at \( x = \pm 1 \) because the denominator \( 1 - x^2 \) becomes zero. Therefore, \( f'(x) \) is not defined at these points.

### Final Answer
Thus, the derivative \( f'(x) \) is:
\[ f'(x) = \frac{2(1 + x^2)}{1 + 2x^2 + x^4}, \quad x \in \mathbb{R}, \, x \neq \pm 1 \]