If `f(x)=|cosx-sinx| `, then `f'(pi/4)` is equal to

Options

A. `sqrt2`
B. `-sqrt2`
C. 0
D. None of these

Answer

To find \( f'\left(\frac{\pi}{4}\right) \) for the function \( f(x) = |\cos x - \sin x| \), we will follow these steps:

### Step 1: Determine the function behavior around \( x = \frac{\pi}{4} \)

The first step is to analyze the expression inside the absolute value: \( \cos x - \sin x \).

At \( x = \frac{\pi}{4} \):
\[ \cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Thus,
\[ \cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right) = 0 \]

### Step 2: Determine the sign of \( \cos x - \sin x \)

Next, we need to determine when \( \cos x - \sin x \) is positive or negative.

- For \( x < \frac{\pi}{4} \):
- \( \cos x > \sin x \) (since cosine starts higher than sine in the first quadrant).

- For \( x > \frac{\pi}{4} \):
- \( \cos x < \sin x \) (since sine starts to rise above cosine after \( \frac{\pi}{4} \)).

### Step 3: Define \( f(x) \) without the absolute value

From the above analysis, we can express \( f(x) \) as:
\[ f(x) = \begin{cases} \cos x - \sin x & \text{if } x < \frac{\pi}{4} \\ -(\cos x - \sin x) & \text{if } x > \frac{\pi}{4} \end{cases} \]

At \( x = \frac{\pi}{4} \), we have \( f\left(\frac{\pi}{4}\right) = 0 \).

### Step 4: Differentiate \( f(x) \)

Now we differentiate \( f(x) \) in both cases:

1. For \( x < \frac{\pi}{4} \):
\[ f'(x) = -\sin x - \cos x \]

2. For \( x > \frac{\pi}{4} \):
\[ f'(x) = -(-\sin x - \cos x) = \sin x + \cos x \]

### Step 5: Evaluate the derivative at \( x = \frac{\pi}{4} \)

To find \( f'\left(\frac{\pi}{4}\right) \), we need to check the left-hand limit (LHL) and right-hand limit (RHL):

- **Left-hand limit** as \( x \to \frac{\pi}{4}^- \):
\[ f'\left(\frac{\pi}{4}^-\right) = -\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} \]

- **Right-hand limit** as \( x \to \frac{\pi}{4}^+ \):
\[ f'\left(\frac{\pi}{4}^+\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \]

### Step 6: Conclusion

Since the left-hand limit and right-hand limit are not equal:
\[ f'\left(\frac{\pi}{4}\right) \text{ does not exist.} \]