Step-by-step Solution:

The function \( f(x) \) has two parts depending on the value of \( x \) because of the absolute value in the denominator: \[\] 1. For \( x > 0 \): \[ f(x) = \frac{x}{1 + x}. \] This is a smooth function, and its derivative is: \[ f'(x) = \frac{1 \cdot (1 + x) - x \cdot 1}{(1 + x)^2} = \frac{1}{(1 + x)^2}. \] The derivative exists for all \( x > 0 \). \[\] 2. For \( x < 0 \): \[ f(x) = \frac{x}{1 - x}. \] Similarly, its derivative is: \[ f'(x) = \frac{1 \cdot (1 - x) + x \cdot 1}{(1 - x)^2} = \frac{1}{(1 - x)^2}. \] The derivative exists for all \( x < 0 \). \[\] 3. At \( x = 0 \): The derivative at \( x = 0 \) requires us to check the limits of the left-hand and right-hand derivatives: - The left-hand derivative (as \( x \to 0^- \)) is: \[ \lim_{x \to 0^-} f'(x) = \lim_{x \to 0^-} \frac{1}{(1 - x)^2} = 1. \] - The right-hand derivative (as \( x \to 0^+ \)) is: \[ \lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} \frac{1}{(1 + x)^2} = 1. \] Since both the left-hand and right-hand derivatives match, the function is differentiable at \( x = 0 \). \[\]Conclusion: The function is differentiable for all real values of \( x \), i.e., \( (-\infty, \infty) \). Thus, the correct answer is: \[ (-\infty, \infty) \]