the derivative of the function ` cot^(-1) "" [(cos 2x ]^(1//2) ] " at" x=(pi)/(6) ` is

Options

A. `(2//3)^(1//2)`
B. `(1//3)^(1//2)`
C. `3^(1//2)`
D. `6^(1//2)`

Answer

To find the derivative of the function \( y = \cot^{-1} \left( \sqrt{\cos(2x)} \right) \) at \( x = \frac{\pi}{6} \), we will follow these steps:

### Step 1: Differentiate the Function
Using the chain rule, we differentiate \( y \) with respect to \( x \).

\[ y = \cot^{-1} \left( \sqrt{\cos(2x)} \right) \]

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

\[ \frac{dy}{du} = -\frac{1}{1 + u^2} \]

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

### Step 2: Differentiate the Inner Function
Now, we need to differentiate \( u = \sqrt{\cos(2x)} \).

Using the chain rule again:

\[ \frac{du}{dx} = \frac{1}{2\sqrt{\cos(2x)}} \cdot \frac{d}{dx}(\cos(2x)) \]

The derivative of \( \cos(2x) \) is:

\[ \frac{d}{dx}(\cos(2x)) = -\sin(2x) \cdot 2 = -2\sin(2x) \]

Thus,

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

### Step 3: Combine Derivatives
Now we can combine the derivatives using the chain rule:

\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\frac{1}{1 + u^2} \cdot \left(-\frac{\sin(2x)}{\sqrt{\cos(2x)}}\right) \]

Substituting \( u = \sqrt{\cos(2x)} \):

\[ \frac{dy}{dx} = \frac{\sin(2x)}{\sqrt{\cos(2x)} \left( 1 + \cos(2x) \right)} \]

### Step 4: Evaluate at \( x = \frac{\pi}{6} \)
Now we need to evaluate this derivative at \( x = \frac{\pi}{6} \).

First, calculate \( \cos(2x) \) at \( x = \frac{\pi}{6} \):

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

Next, calculate \( \sin(2x) \):

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

Now substitute these values into the derivative:

\[ \frac{dy}{dx} = \frac{\frac{\sqrt{3}}{2}}{\sqrt{\frac{1}{2}} \left( 1 + \frac{1}{2} \right)} = \frac{\frac{\sqrt{3}}{2}}{\sqrt{\frac{1}{2}} \cdot \frac{3}{2}} \]

Calculating \( \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \):

\[ \frac{dy}{dx} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}} \cdot \frac{3}{2}} = \frac{\sqrt{3}}{2} \cdot \frac{2\sqrt{2}}{3} = \frac{\sqrt{6}}{3} \]

### Final Answer
Thus, the derivative of the function \( y = \cot^{-1} \left( \sqrt{\cos(2x)} \right) \) at \( x = \frac{\pi}{6} \) is:

\[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{6}} = \frac{\sqrt{6}}{3} \]

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