If `y = cot^(-1) ((1 -x)/(1 +x)) " then " (dy)/(dx) =` ?

Options

A. `(-1)/((1 +x^(2)))`
B. `(1)/((1 + x^(2)))`
C. `(1)/((1 + x^(2))^(.^(3)//_(2)))`
D. none of these

Explanation

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

### Step 1: Rewrite the function
We know that \( \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \). Therefore, we can rewrite \( y \) as:
\[ y = \tan^{-1}\left(\frac{1}{\frac{1 - x}{1 + x}}\right) = \tan^{-1}\left(\frac{1 + x}{1 - x}\right) \]

### Step 2: Set \( x = \tan(\theta) \)
Let \( x = \tan(\theta) \). Then, we can express \( y \) in terms of \( \theta \):
\[ y = \tan^{-1}\left(\frac{1 + \tan(\theta)}{1 - \tan(\theta)}\right) \]

### Step 3: Use the tangent addition formula
Using the tangent addition formula, we have:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
By letting \( a = 45^\circ \) (or \( \frac{\pi}{4} \)) and \( b = \theta \), we get:
\[ y = \tan^{-1}(\tan(45^\circ + \theta)) = 45^\circ + \theta = \frac{\pi}{4} + \theta \]

### Step 4: Express \( \theta \) in terms of \( x \)
Since \( \theta = \tan^{-1}(x) \), we can substitute back:
\[ y = \frac{\pi}{4} + \tan^{-1}(x) \]

### Step 5: Differentiate \( y \)
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 0 + \frac{d}{dx}(\tan^{-1}(x)) \]
Using the derivative of \( \tan^{-1}(x) \):
\[ \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1 + x^2} \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{1}{1 + x^2} \]

### Final Answer
The derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{1}{1 + x^2} \]