Let `f(x)=[|x|-|x-1|]^(2)`
What is f'(x) equal to when `0lt xlt1`?

Options

A. 0
B. `2x-1`
C. `4x-2`
D. `8x-4`

Explanation

To find the derivative of the function \( f(x) = [|x| - |x - 1|]^2 \) for the interval \( 0 < x < 1 \), we will follow these steps:

### Step 1: Simplify the function for the interval \( 0 < x < 1 \)

In the interval \( 0 < x < 1 \):
- \( |x| = x \) (since \( x \) is positive)
- \( |x - 1| = 1 - x \) (since \( x - 1 \) is negative)

Thus, we can rewrite the function:
\[ f(x) = [x - (1 - x)]^2 = [x - 1 + x]^2 = [2x - 1]^2 \]

### Step 2: Differentiate the function

Now we will differentiate \( f(x) = [2x - 1]^2 \) using the chain rule. The chain rule states that if \( y = g(h(x)) \), then \( \frac{dy}{dx} = g'(h(x)) \cdot h'(x) \).

Let \( g(u) = u^2 \) where \( u = 2x - 1 \).

1. Differentiate \( g(u) \):
\[ g'(u) = 2u \]

2. Differentiate \( u \):
\[ h(x) = 2x - 1 \quad \Rightarrow \quad h'(x) = 2 \]

Now applying the chain rule:
\[ f'(x) = g'(h(x)) \cdot h'(x) = 2(2x - 1) \cdot 2 = 4(2x - 1) \]

### Step 3: Final expression for the derivative

Thus, the derivative of \( f(x) \) in the interval \( 0 < x < 1 \) is:
\[ f'(x) = 4(2x - 1) \]

### Summary

The final answer for \( f'(x) \) when \( 0 < x < 1 \) is:
\[ f'(x) = 4(2x - 1) \]

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