If `f(x)=sqrt(x^(2)-2x+1),` then f' (x) ?

Options

A. `f'(x)=1" for all "x`
B. `f'(x)=-1" for all "xle1`
C. `f'(x)=1" for all "xgt1`
D. none of these

Answer

To find the derivative \( f'(x) \) of the function \( f(x) = \sqrt{x^2 - 2x + 1} \), we can follow these steps:

### Step 1: Simplify the function
First, notice that the expression inside the square root can be simplified:
\[ x^2 - 2x + 1 = (x - 1)^2 \]
Thus, we can rewrite the function as:
\[ f(x) = \sqrt{(x - 1)^2} \]
Since the square root of a square gives the absolute value, we have:
\[ f(x) = |x - 1| \]

### Step 2: Differentiate using the chain rule
To differentiate \( f(x) \), we will consider two cases based on the definition of the absolute value.

1. **Case 1**: When \( x - 1 \geq 0 \) (i.e., \( x \geq 1 \)):
\[ f(x) = x - 1 \]
The derivative is:
\[ f'(x) = 1 \]

2. **Case 2**: When \( x - 1 < 0 \) (i.e., \( x < 1 \)):
\[ f(x) = -(x - 1) = 1 - x \]
The derivative is:
\[ f'(x) = -1 \]

### Step 3: Combine the results
Thus, we can express the derivative \( f'(x) \) as:
\[ f'(x) = \begin{cases} 1 & \text{if } x \geq 1 \\ -1 & \text{if } x < 1 \end{cases} \]

### Final Answer
The derivative \( f'(x) \) is:
\[ f'(x) = \begin{cases} 1 & \text{if } x \geq 1 \\ -1 & \text{if } x < 1 \end{cases} \]

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