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

Answer

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

### Step 1: Rewrite the expression
Let \( y = \tan^{-1}\left(\frac{1+x}{1-x}\right) \).

### Step 2: Use a trigonometric identity
We can express the argument of the arctangent in terms of tangent. We know that:
\[ \tan\left(\frac{\pi}{4} + \theta\right) = \frac{\tan\frac{\pi}{4} + \tan\theta}{1 - \tan\frac{\pi}{4} \tan\theta} \]
Here, \( \tan\frac{\pi}{4} = 1 \) and \( \tan\theta = x \). Thus, we can rewrite:
\[ \frac{1+x}{1-x} = \tan\left(\frac{\pi}{4} + \tan^{-1}(x)\right) \]

### Step 3: Simplify the expression
So we can rewrite \( y \):
\[ y = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + \tan^{-1}(x)\right)\right) \]
This simplifies to:
\[ y = \frac{\pi}{4} + \tan^{-1}(x) \]

### Step 4: Differentiate with respect to \( x \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4} + \tan^{-1}(x)\right) \]
Since \( \frac{\pi}{4} \) is a constant, its derivative is 0. The derivative of \( \tan^{-1}(x) \) is:
\[ \frac{d}{dx}\left(\tan^{-1}(x)\right) = \frac{1}{1+x^2} \]
Thus:
\[ \frac{dy}{dx} = 0 + \frac{1}{1+x^2} = \frac{1}{1+x^2} \]

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

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