Question

The domain of the derivative of the function \( f(x) = \dfrac{x}{1 + |x|} \) is

• A. \( [0, \infty) \)
• B. \( (-\infty, 0) \)
• C. \( (-\infty, \infty) \)
• D. \( (0, \infty) \)

Hint:

Always split absolute value functions into piecewise cases before differentiating, and check continuity and derivative limits at the split point.

Answer 1

The Correct Option is C

We are given the function: \[ f(x) = \frac{x}{1 + |x|} \] We need to find the domain of \( f'(x) \), the derivative of this function. Step 1: Understand the behavior of \( f(x) \) The function \( f(x) \) involves an absolute value, so we break it into two cases:
For \( x \geq 0 \), \( |x| = x \Rightarrow f(x) = \dfrac{x}{1 + x} \)
For \( x<0 \), \( |x| = -x \Rightarrow f(x) = \dfrac{x}{1 - x} \) Step 2: Compute the derivative piecewise Case 1: \( x>0 \) \[ f(x) = \frac{x}{1 + x} \Rightarrow f'(x) = \frac{(1 + x)(1) - x(1)}{(1 + x)^2} = \frac{1 + x - x}{(1 + x)^2} = \frac{1}{(1 + x)^2} \] Case 2: \( x<0 \) \[ f(x) = \frac{x}{1 - x} \Rightarrow f'(x) = \frac{(1 - x)(1) - x(-1)}{(1 - x)^2} = \frac{1 - x + x}{(1 - x)^2} = \frac{1}{(1 - x)^2} \] Step 3: Check differentiability at \( x = 0 \) We calculate the left-hand and right-hand derivatives at \( x = 0 \): \[ \lim_{x \to 0^-} f'(x) = \frac{1}{(1 - 0)^2} = 1 \] \[ \lim_{x \to 0^+} f'(x) = \frac{1}{(1 + 0)^2} = 1 \] Since both side limits exist and are equal, and since \( f(x) \) is continuous at \( x = 0 \), the derivative exists at \( x = 0 \). Step 4: Conclusion So the derivative exists for all real \( x \). Hence, the domain of the derivative is: \[ \boxed{(-\infty, \infty)} \]