Differentiate the following w.r.t. x :
`cot^(-1)((1+x)/(1-x))`

Answer

To differentiate the function \( y = \cot^{-1}\left(\frac{1+x}{1-x}\right) \) with respect to \( x \), we can use the chain rule and the quotient rule of differentiation. Here’s a step-by-step solution:

### Step 1: Define the function
Let:
\[ y = \cot^{-1}(u) \]
where
\[ u = \frac{1+x}{1-x} \]

### Step 2: Differentiate \( y \) with respect to \( u \)
Using the derivative of the inverse cotangent function:
\[ \frac{dy}{du} = -\frac{1}{1 + u^2} \]

### Step 3: Differentiate \( u \) with respect to \( x \)
Using the quotient rule, where \( u = \frac{a}{b} \) with \( a = 1+x \) and \( b = 1-x \):
\[ \frac{du}{dx} = \frac{b \frac{da}{dx} - a \frac{db}{dx}}{b^2} \]
Calculating \( \frac{da}{dx} \) and \( \frac{db}{dx} \):
- \( \frac{da}{dx} = 1 \)
- \( \frac{db}{dx} = -1 \)

Now substituting into the quotient rule:
\[ \frac{du}{dx} = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2} = \frac{(1-x) + (1+x)}{(1-x)^2} = \frac{2}{(1-x)^2} \]

### Step 4: Apply the chain rule
Now we can use the chain rule to find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\frac{1}{1 + u^2} \cdot \frac{2}{(1-x)^2} \]

### Step 5: Substitute \( u \) back into the equation
Now we need to express \( 1 + u^2 \):
\[ u = \frac{1+x}{1-x} \implies u^2 = \left(\frac{1+x}{1-x}\right)^2 = \frac{(1+x)^2}{(1-x)^2} \]
Thus,
\[ 1 + u^2 = 1 + \frac{(1+x)^2}{(1-x)^2} = \frac{(1-x)^2 + (1+x)^2}{(1-x)^2} \]
Calculating the numerator:
\[ (1-x)^2 + (1+x)^2 = (1 - 2x + x^2) + (1 + 2x + x^2) = 2 + 2x^2 \]
So,
\[ 1 + u^2 = \frac{2 + 2x^2}{(1-x)^2} = \frac{2(1+x^2)}{(1-x)^2} \]

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

### Final Result
Thus, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = -\frac{(1-x)^2}{1+x^2} \]

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