If `f(x)=|cosx-sinx|`, then `f'(pi/2)` is equal to

Options

A. 1
B. -1
C. 0
D. none of these

Explanation

To find \( f'( \frac{\pi}{2} ) \) for the function \( f(x) = | \cos x - \sin x | \), we need to analyze the function and its derivative step by step.

### Step 1: Determine the expression for \( f(x) \)

The function is defined as:
\[ f(x) = | \cos x - \sin x | \]

### Step 2: Identify where \( \cos x - \sin x \) changes sign

To find the points where the expression inside the absolute value changes sign, we set:
\[ \cos x - \sin x = 0 \]
This simplifies to:
\[ \cos x = \sin x \]
This occurs at:
\[ x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 3: Analyze the intervals

We will analyze the intervals around \( x = \frac{\pi}{4} \):
- For \( 0 \leq x < \frac{\pi}{4} \), \( \cos x > \sin x \), so:
\[ f(x) = \cos x - \sin x \]
- For \( \frac{\pi}{4} < x < \frac{5\pi}{4} \), \( \sin x > \cos x \), so:
\[ f(x) = \sin x - \cos x \]
- For \( \frac{5\pi}{4} < x < 2\pi \), \( \cos x > \sin x \), so:
\[ f(x) = \cos x - \sin x \]

### Step 4: Find the derivative \( f'(x) \)

Now we differentiate \( f(x) \) in each interval:
- For \( 0 \leq x < \frac{\pi}{4} \):
\[ f'(x) = -\sin x - \cos x \]
- For \( \frac{\pi}{4} < x < \frac{5\pi}{4} \):
\[ f'(x) = \cos x + \sin x \]
- For \( \frac{5\pi}{4} < x < 2\pi \):
\[ f'(x) = -\sin x - \cos x \]

### Step 5: Evaluate \( f'(\frac{\pi}{2}) \)

Since \( \frac{\pi}{2} \) is in the interval \( \left( \frac{\pi}{4}, \frac{5\pi}{4} \right) \), we use:
\[ f'(x) = \cos x + \sin x \]
Now substituting \( x = \frac{\pi}{2} \):
\[ f'(\frac{\pi}{2}) = \cos\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = 0 + 1 = 1 \]

### Final Answer

Thus, the value of \( f'(\frac{\pi}{2}) \) is:
\[ \boxed{1} \]