The derivative of y=|x-2| at x=2 is

Options

A. 1
B. `-1`
C. 0
D. None of these

Answer

To find the derivative of the function \( y = |x - 2| \) at the point \( x = 2 \), we will follow these steps:

### Step 1: Understand the function
The function \( y = |x - 2| \) is a piecewise function. We need to express it in a form that we can differentiate. The absolute value function can be broken down as follows:
- For \( x < 2 \), \( |x - 2| = -(x - 2) = 2 - x \)
- For \( x \geq 2 \), \( |x - 2| = x - 2 \)

### Step 2: Identify the critical point
The critical point occurs at \( x = 2 \) because the expression inside the absolute value changes sign at this point.

### Step 3: Find the derivative from the left and right
To determine if the derivative exists at \( x = 2 \), we will calculate the left-hand derivative and the right-hand derivative.

#### Left-hand derivative (as \( x \) approaches 2 from the left):
For \( x < 2 \):
\[ y = 2 - x \]
Taking the derivative:
\[ \frac{dy}{dx} = -1 \]
Thus, the left-hand derivative at \( x = 2 \) is:
\[ \lim_{h \to 0^-} \frac{f(2 + h) - f(2)}{h} = -1 \]

#### Right-hand derivative (as \( x \) approaches 2 from the right):
For \( x \geq 2 \):
\[ y = x - 2 \]
Taking the derivative:
\[ \frac{dy}{dx} = 1 \]
Thus, the right-hand derivative at \( x = 2 \) is:
\[ \lim_{h \to 0^+} \frac{f(2 + h) - f(2)}{h} = 1 \]

### Step 4: Compare the left-hand and right-hand derivatives
Now we compare the left-hand and right-hand derivatives:
- Left-hand derivative at \( x = 2 \): \( -1 \)
- Right-hand derivative at \( x = 2 \): \( 1 \)

Since the left-hand derivative and right-hand derivative are not equal, the derivative of \( y = |x - 2| \) at \( x = 2 \) does not exist.

### Conclusion
The derivative of \( y = |x - 2| \) at \( x = 2 \) is **undefined**.

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