Find the derivative of `tan^(-1)(2x)/(1-x^2)wdotrdottsin^(-1)(2x)/(1+x^2)`

Explanation

To find the derivative of \( y = \frac{\tan^{-1}(2x)}{1 - x^2} \) with respect to \( t = \frac{\sin^{-1}(2x)}{1 + x^2} \), we will follow these steps:

### Step 1: Define the Variables
Let:
- \( y = \frac{\tan^{-1}(2x)}{1 - x^2} \)
- \( t = \frac{\sin^{-1}(2x)}{1 + x^2} \)

### Step 2: Change of Variables
We can express \( x \) in terms of \( \theta \) by letting \( x = \tan(\theta) \). Then we can express \( y \) and \( t \) in terms of \( \theta \).

### Step 3: Calculate \( y \)
Using the identity for the tangent of double angles:
\[ y = \tan^{-1}(2 \tan(\theta)) = \tan^{-1}\left(\frac{2 \tan(\theta)}{1 - \tan^2(\theta)}\right) = 2\theta \]
Thus, we have:
\[ y = 2\theta \]

### Step 4: Calculate \( t \)
Now, substituting \( x = \tan(\theta) \) into \( t \):
\[ t = \sin^{-1}\left(\frac{2 \tan(\theta)}{1 + \tan^2(\theta)}\right) = \sin^{-1}(\sin(2\theta)) = 2\theta \]
Thus, we have:
\[ t = 2\theta \]

### Step 5: Relate \( y \) and \( t \)
From the above, we see that:
\[ y = t \]

### Step 6: Differentiate
Now, we differentiate both sides with respect to \( x \):
\[ \frac{dy}{dx} = \frac{dt}{dx} \]

### Step 7: Find \( \frac{dy}{dt} \)
Since \( y = t \), we have:
\[ \frac{dy}{dt} = 1 \]

### Final Result
Thus, the derivative \( \frac{dy}{dt} \) is:
\[ \frac{dy}{dt} = 1 \]