If `f(x)=|sinx-|cosx||,` then the value of `f^(')(x)` at `x=(7pi)/6` is

Options

A. `(sqrt3+1)/(2)`
B. `(1-sqrt3)/(2)`
C. `(sqrt3-1)/(2)`
D. `(-1-sqrt3)/(2)`

Answer

To find the value of \( f'(x) \) at \( x = \frac{7\pi}{6} \) for the function \( f(x) = |\sin x - |\cos x|| \), we will follow these steps:

### Step 1: Determine the Quadrant
The angle \( \frac{7\pi}{6} \) lies in the third quadrant. In this quadrant, both sine and cosine are negative.

### Step 2: Simplify the Function
Since \( \frac{7\pi}{6} \) is in the third quadrant:
- \( \sin\left(\frac{7\pi}{6}\right) < 0 \) so \( |\sin\left(\frac{7\pi}{6}\right)| = -\sin\left(\frac{7\pi}{6}\right) \)
- \( \cos\left(\frac{7\pi}{6}\right) < 0 \) so \( |\cos\left(\frac{7\pi}{6}\right)| = -\cos\left(\frac{7\pi}{6}\right) \)

Thus, we can write:
\[ f(x) = -\sin x + \cos x \]

### Step 3: Differentiate the Function
Now we differentiate \( f(x) \):
\[ f'(x) = -\cos x - \sin x \]

### Step 4: Evaluate the Derivative at \( x = \frac{7\pi}{6} \)
Now we need to evaluate \( f'\left(\frac{7\pi}{6}\right) \):
\[ f'\left(\frac{7\pi}{6}\right) = -\cos\left(\frac{7\pi}{6}\right) - \sin\left(\frac{7\pi}{6}\right) \]

### Step 5: Calculate the Values of Sine and Cosine
Using the angle \( \frac{7\pi}{6} = \pi + \frac{\pi}{6} \):
- \( \sin\left(\frac{7\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2} \)
- \( \cos\left(\frac{7\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} \)

Substituting these values into the derivative:
\[ f'\left(\frac{7\pi}{6}\right) = -\left(-\frac{\sqrt{3}}{2}\right) - \left(-\frac{1}{2}\right) \]
\[ = \frac{\sqrt{3}}{2} + \frac{1}{2} \]
\[ = \frac{\sqrt{3} + 1}{2} \]

### Final Answer
Thus, the value of \( f'\left(\frac{7\pi}{6}\right) \) is:
\[ \frac{\sqrt{3} + 1}{2} \]

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