If `f(x)=cot^(-1)sqrt(cos2x)," then: "f'((pi)/(6))=`

Options

A. `(1)/sqrt(3)`
B. `(2)/sqrt(3)`
C. `(sqrt2)/(3)`
D. `-(2)/(sqrt3)`

Answer

To solve the problem, we need to find the derivative of the function \( f(x) = \cot^{-1}(\sqrt{\cos(2x)}) \) and then evaluate it at \( x = \frac{\pi}{6} \).

### Step 1: Differentiate the function

The derivative of \( \cot^{-1}(u) \) is given by:

\[ f'(x) = -\frac{u'}{1 + u^2} \]

where \( u = \sqrt{\cos(2x)} \).

First, we need to find \( u' \):

\[ u = \sqrt{\cos(2x)} \]

Using the chain rule, we differentiate \( u \):

\[ u' = \frac{1}{2\sqrt{\cos(2x)}} \cdot (-\sin(2x) \cdot 2) = -\frac{\sin(2x)}{\sqrt{\cos(2x)}} \]

Now substituting \( u \) and \( u' \) into the derivative formula:

\[ f'(x) = -\frac{-\frac{\sin(2x)}{\sqrt{\cos(2x)}}}{1 + \cos(2x)} \]

This simplifies to:

\[ f'(x) = \frac{\sin(2x)}{\sqrt{\cos(2x)}(1 + \cos(2x))} \]

### Step 2: Evaluate the derivative at \( x = \frac{\pi}{6} \)

Now we substitute \( x = \frac{\pi}{6} \):

1. Calculate \( \cos(2x) \):

\[ \cos(2 \cdot \frac{\pi}{6}) = \cos(\frac{\pi}{3}) = \frac{1}{2} \]

2. Calculate \( \sin(2x) \):

\[ \sin(2 \cdot \frac{\pi}{6}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \]

3. Substitute these values into \( f'(\frac{\pi}{6}) \):

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

4. Simplify the expression:

\[ = \frac{\frac{\sqrt{3}}{2}}{\sqrt{\frac{1}{2}} \cdot \frac{3}{2}} = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2} \cdot \frac{3}{2}} = \frac{\sqrt{3}}{3\sqrt{2}} \]

5. Finally, simplify:

\[ = \frac{\sqrt{3}}{3\sqrt{2}} = \frac{\sqrt{6}}{6} \]

Thus, the final answer is:

\[ f'\left(\frac{\pi}{6}\right) = \frac{\sqrt{6}}{6} \]