`tan^-1((1+x)/(1-x))`

Answer

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

### Step 1: Set up the function
Let
\[ y = \tan^{-1}\left(\frac{1+x}{1-x}\right) \]

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

### Step 3: Differentiate the inner function
Now we need to differentiate the inner function \( \frac{1+x}{1-x} \) using the quotient rule. The quotient rule states that if \( u = 1+x \) and \( v = 1-x \), then:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Calculating \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
- \( \frac{du}{dx} = 1 \)
- \( \frac{dv}{dx} = -1 \)

Now applying the quotient rule:
\[ \frac{d}{dx}\left(\frac{1+x}{1-x}\right) = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2} = \frac{1 - x + 1 + x}{(1-x)^2} = \frac{2}{(1-x)^2} \]

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

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

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

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