If `f(x)=x/(1+|x|)` for ` x in R`, then `f'(0) =`

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find \( f'(0) \) for the function \( f(x) = \frac{x}{1 + |x|} \), we will analyze the function based on the definition of the absolute value and then differentiate it.

### Step 1: Define the function based on cases
The absolute value function \( |x| \) can be defined in two cases:
- For \( x \geq 0 \), \( |x| = x \)
- For \( x < 0 \), \( |x| = -x \)

Thus, we can rewrite \( f(x) \) as follows:
- If \( x \geq 0 \):
\[ f(x) = \frac{x}{1 + x} \]
- If \( x < 0 \):
\[ f(x) = \frac{x}{1 - x} \]

### Step 2: Find the derivative for each case
Now, we will differentiate \( f(x) \) for both cases.

**Case 1: \( x \geq 0 \)**
\[ f(x) = \frac{x}{1 + x} \]
Using the quotient rule:
\[ f'(x) = \frac{(1 + x)(1) - x(1)}{(1 + x)^2} = \frac{1 + x - x}{(1 + x)^2} = \frac{1}{(1 + x)^2} \]

**Case 2: \( x < 0 \)**
\[ f(x) = \frac{x}{1 - x} \]
Again, using the quotient rule:
\[ f'(x) = \frac{(1 - x)(1) - x(-1)}{(1 - x)^2} = \frac{1 - x + x}{(1 - x)^2} = \frac{1}{(1 - x)^2} \]

### Step 3: Evaluate \( f'(0) \)
Since \( f'(x) \) is defined differently for \( x \geq 0 \) and \( x < 0 \), we need to evaluate \( f'(0) \) using the case for \( x \geq 0 \):
\[ f'(0) = \frac{1}{(1 + 0)^2} = \frac{1}{1^2} = 1 \]

### Final Answer
Thus, \( f'(0) = 1 \).

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