If `f (x) =|sin x-|cos x ||,` then `f '((7pi)/(6))=`

Options

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

Answer

To find \( f' \left( \frac{7\pi}{6} \right) \) for the function \( f(x) = |\sin x - |\cos x|| \), we will go through the following steps:

### Step 1: Evaluate \( |\cos x| \) near \( x = \frac{7\pi}{6} \)

First, we need to evaluate \( |\cos x| \) at \( x = \frac{7\pi}{6} \).

\[ \cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2} \]
Thus,
\[ |\cos\left(\frac{7\pi}{6}\right)| = \frac{\sqrt{3}}{2} \]

### Step 2: Substitute into \( f(x) \)

Now, we can substitute \( |\cos x| \) into the function \( f(x) \):

\[ f(x) = |\sin x - |\cos x|| = |\sin x + \frac{\sqrt{3}}{2}| \]

### Step 3: Evaluate \( \sin x \) near \( x = \frac{7\pi}{6} \)

Next, we evaluate \( \sin x \) at \( x = \frac{7\pi}{6} \):

\[ \sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2} \]

### Step 4: Substitute into \( f(x) \)

Now substituting \( \sin\left(\frac{7\pi}{6}\right) \) into \( f(x) \):

\[ f\left(\frac{7\pi}{6}\right) = \left| -\frac{1}{2} + \frac{\sqrt{3}}{2} \right| = \left| \frac{\sqrt{3}}{2} - \frac{1}{2} \right| = \left| \frac{\sqrt{3} - 1}{2} \right| \]

Since \( \sqrt{3} > 1 \), we have:

\[ f\left(\frac{7\pi}{6}\right) = \frac{\sqrt{3} - 1}{2} \]

### Step 5: Differentiate \( f(x) \)

Now we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx} \left( |\sin x + \frac{\sqrt{3}}{2}| \right) \]

Since \( \sin x + \frac{\sqrt{3}}{2} \) is negative in the neighborhood of \( x = \frac{7\pi}{6} \), we have:

\[ f(x) = -\left( \sin x + \frac{\sqrt{3}}{2} \right) \]

Thus, differentiating gives:

\[ f'(x) = -\cos x \]

### Step 6: Evaluate \( f' \left( \frac{7\pi}{6} \right) \)

Now we can evaluate \( f' \left( \frac{7\pi}{6} \right) \):

\[ f' \left( \frac{7\pi}{6} \right) = -\cos\left(\frac{7\pi}{6}\right) = -\left(-\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2} \]

### Step 7: Final Calculation

Now we need to calculate:

\[ f' \left( \frac{7\pi}{6} \right) = \sin\left(\frac{7\pi}{6}\right) - \cos\left(\frac{7\pi}{6}\right) \]

Substituting the values:

\[ f' \left( \frac{7\pi}{6} \right) = -\frac{1}{2} - \left(-\frac{\sqrt{3}}{2}\right) = -\frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{\sqrt{3} - 1}{2} \]

Thus, the final answer is:

\[ f' \left( \frac{7\pi}{6} \right) = \frac{\sqrt{3} - 1}{2} \]