If `2y=(cot^(-1) ((sqrt3cosx+sinx)/(cosx-sqrt3sinx)))^2, x in (0,pi/2)`then y' is equal to

Options

A. (a)`2x-pi/3`
B. (b)`pi/6-x`
C. (c)`x-pi/6`
D. (d)`pi/3-x`

Explanation

To solve the given problem step by step, we start with the equation provided:

Given:
\[ 2y = \left( \cot^{-1} \left( \frac{\sqrt{3} \cos x + \sin x}{\cos x - \sqrt{3} \sin x} \right) \right)^2 \]

### Step 1: Simplifying the Argument of Cotangent
We can rewrite the expression inside the cotangent inverse using the tangent addition formula. Notice that:

\[ \frac{\sqrt{3} \cos x + \sin x}{\cos x - \sqrt{3} \sin x} = \frac{\tan\left(\frac{\pi}{3}\right) \cos x + \sin x}{\cos x - \tan\left(\frac{\pi}{3}\right) \sin x} \]

This resembles the tangent addition formula:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
where \( a = \frac{\pi}{3} \) and \( b = x \).

Thus, we can express:
\[ \cot^{-1} \left( \frac{\sqrt{3} \cos x + \sin x}{\cos x - \sqrt{3} \sin x} \right) = \frac{\pi}{2} - \tan^{-1} \left( \tan\left(\frac{\pi}{3} + x\right) \right) = \frac{\pi}{3} + x \]

### Step 2: Substitute Back into the Equation
Now substituting back into the equation for \( 2y \):
\[ 2y = \left( \frac{\pi}{3} + x \right)^2 \]

### Step 3: Expand the Square
Expanding the square:
\[ 2y = \left( \frac{\pi}{3} + x \right)^2 = \left( \frac{\pi^2}{9} + \frac{2\pi}{3}x + x^2 \right) \]

### Step 4: Solve for y
Now, divide both sides by 2:
\[ y = \frac{\pi^2}{18} + \frac{\pi}{3}x + \frac{x^2}{2} \]

### Step 5: Differentiate y with Respect to x
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi^2}{18} + \frac{\pi}{3}x + \frac{x^2}{2} \right) \]
Calculating the derivative:
\[ \frac{dy}{dx} = 0 + \frac{\pi}{3} + x \]

### Final Result
Thus, the derivative \( y' \) is:
\[ y' = \frac{\pi}{3} + x \]