Differentiate the following w.r.t. x :
`tan^(-1)((2x)/(1-x^(2))),0ltxlt1`

Answer

To differentiate the function \( y = \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \) with respect to \( x \), we can follow these steps:

### Step 1: Recognize the Form
We start with the function:
\[ y = \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \]
We can use the identity for the tangent of a double angle:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
This suggests that if we let \( x = \tan(\theta) \), then:
\[ \frac{2x}{1 - x^2} = \tan(2\theta) \]
Thus, we can rewrite \( y \) as:
\[ y = \tan^{-1}(\tan(2\theta)) = 2\theta \]

### Step 2: Express \( \theta \) in Terms of \( x \)
Since \( x = \tan(\theta) \), we have:
\[ \theta = \tan^{-1}(x) \]
Substituting this back into our expression for \( y \):
\[ y = 2 \tan^{-1}(x) \]

### Step 3: Differentiate \( y \) with Respect to \( x \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}(x)) \]
Using the derivative of \( \tan^{-1}(x) \):
\[ \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1 + x^2} \]
Thus:
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{1 + x^2} \]

### Step 4: Final Result
Therefore, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{2}{1 + x^2} \]