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 find 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: Identify the intervals for \( f(x) \)

The function \( f(x) \) is defined in two cases based on the value of \( |x| \):
1. For \( |x| \geq 1 \): This means \( x \leq -1 \) or \( x \geq 1 \).
2. For \( |x| < 1 \): This means \( -1 < x < 1 \).

### Step 2: Write \( f(x) \) without absolute values

Now, we can express \( f(x) \) without absolute values:
- For \( x \leq -1 \):
\[ f(x) = \frac{x}{1 + (-x)} = \frac{x}{1 - x} \]
- For \( x \geq 1 \):
\[ f(x) = \frac{x}{1 + x} \]
- For \( -1 < x < 1 \):
\[ f(x) = \frac{x}{1 - x} \]

### Step 3: Differentiate \( f(x) \)

Next, we differentiate \( f(x) \) in each of the intervals:

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

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

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

### Step 4: Determine the domain of \( f'(x) \)

The function \( f'(x) \) is defined as long as the denominators are not zero.

- For \( x \leq -1 \): The expression \( (1 - x)^2 \) is never zero for \( x \leq -1 \).
- For \( -1 < x < 1 \): The expression \( (1 - x)^2 \) is never zero for \( -1 < x < 1 \).
- For \( x \geq 1 \): The expression \( (1 + x)^2 \) is never zero for \( x \geq 1 \).

However, we need to check the points where the original function \( f(x) \) is not defined, specifically at \( x = -1 \) and \( x = 1 \):
- At \( x = -1 \): \( f(-1) \) is defined, but \( f'(-1) \) leads to a division by zero.
- At \( x = 1 \): \( f(1) \) is defined, but \( f'(1) \) leads to a division by zero.

### Conclusion

Thus, the domain of \( f'(x) \) is all real numbers except \( -1 \) and \( 1 \):

\[ \text{Domain of } f'(x) = (-\infty, -1) \cup (-1, 1) \cup (1, \infty) \]