`(d)/(dx)[cot^(-1)((1-x)/(1+x))]=`

Options

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

Answer

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

### Step 1: Use the differentiation formula for cotangent inverse
The differentiation formula for \( \cot^{-1}(x) \) is:
\[ \frac{d}{dx} \left( \cot^{-1}(x) \right) = -\frac{1}{1 + x^2} \]
We will apply this formula to our function where \( x = \frac{1-x}{1+x} \).

### Step 2: Differentiate using the chain rule
Using the chain rule, we have:
\[ \frac{dy}{dx} = -\frac{1}{1 + \left(\frac{1-x}{1+x}\right)^2} \cdot \frac{d}{dx}\left(\frac{1-x}{1+x}\right) \]

### Step 3: Differentiate the inner function
Now we need to differentiate \( \frac{1-x}{1+x} \):
Using the quotient rule:
\[ \frac{d}{dx}\left(\frac{1-x}{1+x}\right) = \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2} \]
Simplifying the numerator:
\[ = \frac{-1 - x - 1 + x}{(1+x)^2} = \frac{-2}{(1+x)^2} \]

### Step 4: Substitute back into the derivative
Now substituting back into our derivative:
\[ \frac{dy}{dx} = -\frac{1}{1 + \left(\frac{1-x}{1+x}\right)^2} \cdot \left(-\frac{2}{(1+x)^2}\right) \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{2}{(1+x)^2 \left(1 + \left(\frac{1-x}{1+x}\right)^2\right)} \]

### Step 5: Simplify the expression
Now we need to simplify \( 1 + \left(\frac{1-x}{1+x}\right)^2 \):
\[ 1 + \left(\frac{1-x}{1+x}\right)^2 = 1 + \frac{(1-x)^2}{(1+x)^2} = \frac{(1+x)^2 + (1-x)^2}{(1+x)^2} \]
Calculating the numerator:
\[ (1+x)^2 + (1-x)^2 = 1 + 2x + x^2 + 1 - 2x + x^2 = 2 + 2x^2 \]
Thus,
\[ 1 + \left(\frac{1-x}{1+x}\right)^2 = \frac{2 + 2x^2}{(1+x)^2} \]

### Step 6: Substitute back into the derivative
Now substituting this back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{2}{(1+x)^2} \cdot \frac{(1+x)^2}{2 + 2x^2} = \frac{2}{2 + 2x^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{1}{1 + x^2} \]

### Final Answer
Thus, the derivative of \( y = \cot^{-1}\left(\frac{1-x}{1+x}\right) \) is:
\[ \frac{dy}{dx} = \frac{1}{1 + x^2} \]