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

Options

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

Answer

To find \( f'(\frac{\pi}{2}) \) for the function \( f(x) = |\cos x - \sin x| \), we will follow these steps:

### Step 1: Determine the expression for \( f(x) \)
The function is given as:
\[ f(x) = |\cos x - \sin x| \]
To analyze this function, we need to determine when \( \cos x - \sin x \) is positive or negative.

### Step 2: Find the points where \( \cos x = \sin x \)
Setting \( \cos x = \sin x \), we can solve for \( x \):
\[ \tan x = 1 \implies x = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z} \]
The first two points in the interval \( [0, 2\pi] \) are \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).

### Step 3: Analyze the intervals
- For \( x \in \left(0, \frac{\pi}{4}\right) \): \( \cos x > \sin x \) so \( f(x) = \cos x - \sin x \)
- For \( x \in \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \): \( \sin x > \cos x \) so \( f(x) = -(\cos x - \sin x) = \sin x - \cos x \)
- For \( x \in \left(\frac{5\pi}{4}, 2\pi\right) \): \( \cos x > \sin x \) so \( f(x) = \cos x - \sin x \)

### Step 4: Find \( f(x) \) in the relevant interval
Since \( \frac{\pi}{2} \) is in the interval \( \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \), we use:
\[ f(x) = \sin x - \cos x \]

### Step 5: Differentiate \( f(x) \)
Now we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\sin x - \cos x) = \cos x + \sin x \]

### Step 6: Evaluate \( f'(\frac{\pi}{2}) \)
Now we substitute \( x = \frac{\pi}{2} \):
\[ f'(\frac{\pi}{2}) = \cos\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = 0 + 1 = 1 \]

### Final Answer
Thus, \( f'(\frac{\pi}{2}) = 1 \).

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