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

Options

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

Explanation

To find \( f'(0) \) for the function \( f(x) = \frac{x}{1 + |x|} \), we need to consider the behavior of the function around \( x = 0 \). Since the absolute value function can change the expression of \( f(x) \) depending on whether \( x \) is positive or negative, we will analyze both cases.

### Step-by-Step Solution

1. **Define the function based on the sign of \( x \)**:
\[ f(x) = \begin{cases} \frac{x}{1 + x} & \text{if } x \geq 0 \\ \frac{x}{1 - x} & \text{if } x < 0 \end{cases} \]

2. **Calculate the left-hand derivative (LHD) at \( x = 0 \)**:
\[ \text{LHD} = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{f(h)}{h} \]
For \( h < 0 \), we use \( f(h) = \frac{h}{1 - h} \):
\[ \text{LHD} = \lim_{h \to 0^-} \frac{\frac{h}{1 - h}}{h} = \lim_{h \to 0^-} \frac{1}{1 - h} = \frac{1}{1 - 0} = 1 \]

3. **Calculate the right-hand derivative (RHD) at \( x = 0 \)**:
\[ \text{RHD} = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{f(h)}{h} \]
For \( h \geq 0 \), we use \( f(h) = \frac{h}{1 + h} \):
\[ \text{RHD} = \lim_{h \to 0^+} \frac{\frac{h}{1 + h}}{h} = \lim_{h \to 0^+} \frac{1}{1 + h} = \frac{1}{1 + 0} = 1 \]

4. **Conclusion**:
Since both the left-hand derivative and the right-hand derivative at \( x = 0 \) are equal:
\[ f'(0) = \text{LHD} = \text{RHD} = 1 \]

Thus, the value of \( f'(0) \) is \( \boxed{1} \).