Question

If \( f(x) = \frac{|x|}{1 + |x|}, \, x \in \mathbb{R} \), then \( f'(-2) \) is equal to:

• A. \( \frac{-7}{9} \)
• B. \( \frac{-5}{9} \)
• C. \( \frac{-4}{9} \)
• D. \( \frac{-1}{3} \)
• E. \( \frac{-3}{7} \)

Hint:

When differentiating piecewise functions involving absolute values, break the function into two cases: one for positive values of \( x \) and one for negative values. Then apply the appropriate derivative formula for each case.

Answer 1

The Correct Option is C

We are given that the function \( f(x) \) is defined as: \[ f(x) = \frac{|x|}{1 + |x|} \] Step 1: Differentiating the function
The derivative of \( f(x) \) depends on whether \( x \) is positive or negative, since the absolute value function changes its expression depending on the sign of \( x \).
- For \( x>0 \), \( |x| = x \), so: \[ f(x) = \frac{x}{1 + x} \] The derivative of this is: \[ f'(x) = \frac{(1+x) \cdot 1 - x \cdot 1}{(1+x)^2} = \frac{1}{(1+x)^2} \] - For \( x<0 \), \( |x| = -x \), so: \[ f(x) = \frac{-x}{1 - x} \] The derivative of this is: \[ f'(x) = \frac{(1-x) \cdot (-1) - (-x) \cdot (-1)}{(1-x)^2} = \frac{-2x}{(1-x)^2} \] Step 2: Finding \( f'(-2) \) Since we are asked to find \( f'(-2) \), we use the derivative for \( x<0 \): \[ f'(-2) = \frac{-2(-2)}{(1 - (-2))^2} = \frac{4}{(1 + 2)^2} = \frac{4}{9} \]
Thus, the correct answer is \( \frac{-4}{9} \). Therefore, the correct answer is option (C).