Let `f(x)= {{:((x)/(1+|x|)",", |x| ge1), ((x)/(1-|x|)",", |x| lt 1):},` then domain of `f'(x)` is:

Options

A. `(-oo,oo)`
B. `(-oo,oo)-{-1,0,1}`
C. `(-oo,oo)-{-1,1}`
D. `(-oo,oo) -{0}`

Answer

To determine the domain of \( f'(x) \) for the given piecewise function

\[ f(x) = \begin{cases} \frac{x}{1 + |x|} & \text{if } |x| \geq 1 \\ \frac{x}{1 - |x|} & \text{if } |x| < 1 \end{cases} \]

we will follow these steps:

### Step 1: Define the function for different intervals
We can rewrite the function \( f(x) \) based on the absolute value conditions:

1. For \( x \geq 1 \): \( f(x) = \frac{x}{1 + x} \)
2. For \( -1 < x < 1 \): \( f(x) = \frac{x}{1 - x} \)
3. For \( x \leq -1 \): \( f(x) = \frac{x}{1 - x} \)

### Step 2: Differentiate \( f(x) \)
Now we will differentiate \( f(x) \) in each of these intervals.

1. For \( x \geq 1 \):
\[ f'(x) = \frac{(1 + x) \cdot 1 - x \cdot 1}{(1 + x)^2} = \frac{1}{(1 + x)^2} \]

2. For \( -1 < x < 1 \):
\[ f'(x) = \frac{(1 - x) \cdot 1 - x \cdot (-1)}{(1 - x)^2} = \frac{1}{(1 - x)^2} \]

3. For \( x \leq -1 \):
\[ f'(x) = \frac{(1 - x) \cdot 1 - x \cdot (-1)}{(1 - x)^2} = \frac{1}{(1 - x)^2} \]

### Step 3: Check for continuity and differentiability at critical points
We need to check the points where the definition of \( f(x) \) changes, which are \( x = -1 \), \( x = 0 \), and \( x = 1 \).

- **At \( x = -1 \)**:
- Left-hand derivative (LHD) as \( x \to -1^- \):
\[ f'(-1) = \frac{1}{(1 - (-1))^2} = \frac{1}{4} \]
- Right-hand derivative (RHD) as \( x \to -1^+ \):
\[ f'(-1) = \frac{1}{(1 + (-1))^2} = \frac{1}{0} \text{ (undefined)} \]
- Since LHD and RHD are not equal, \( f'(x) \) is not defined at \( x = -1 \).

- **At \( x = 0 \)**:
- LHD as \( x \to 0^- \):
\[ f'(0) = \frac{1}{(1 - 0)^2} = 1 \]
- RHD as \( x \to 0^+ \):
\[ f'(0) = \frac{1}{(1 + 0)^2} = 1 \]
- Since LHD and RHD are equal, \( f'(x) \) is defined at \( x = 0 \).

- **At \( x = 1 \)**:
- LHD as \( x \to 1^- \):
\[ f'(1) = \frac{1}{(1 - 1)^2} = \frac{1}{0} \text{ (undefined)} \]
- RHD as \( x \to 1^+ \):
\[ f'(1) = \frac{1}{(1 + 1)^2} = \frac{1}{4} \]
- Since LHD and RHD are not equal, \( f'(x) \) is not defined at \( x = 1 \).

### Step 4: Determine the domain of \( f'(x) \)
From the analysis, we find that \( f'(x) \) is defined for:
- \( (-\infty, -1) \)
- \( (-1, 1) \)
- \( (1, \infty) \)

Thus, the domain of \( f'(x) \) is:

\[ (-\infty, -1) \cup (-1, 1) \cup (1, \infty) \]

### Final Answer
The domain of \( f'(x) \) is \( (-\infty, -1) \cup (-1, 1) \cup (1, \infty) \).