Differentiate `tan^(-1)((2x)/(1-x^(2)))` w.r.t. `sin^(-1)((2x)/(1+x^(2))).`

Answer

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

### Step 1: Define the Functions
Let:
- \( U = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \)
- \( V = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \)

### Step 2: Use Trigonometric Identities
We can use the trigonometric identities to simplify \( U \) and \( V \):
- From the double angle formulas, we know:
\[ \frac{2\tan(\theta)}{1-\tan^2(\theta)} = \tan(2\theta) \]
Thus, if we let \( x = \tan(\theta) \), we have:
\[ U = \tan^{-1}(\tan(2\theta)) = 2\theta \]

- Similarly, for \( V \):
\[ \frac{2\tan(\theta)}{1+\tan^2(\theta)} = \sin(2\theta) \]
Therefore:
\[ V = \sin^{-1}(\sin(2\theta)) = 2\theta \]

### Step 3: Relate \( \theta \) to \( x \)
Since \( x = \tan(\theta) \), we can express \( \theta \) as:
\[ \theta = \tan^{-1}(x) \]
Thus:
\[ U = 2\tan^{-1}(x) \quad \text{and} \quad V = 2\tan^{-1}(x) \]

### Step 4: Differentiate \( U \) and \( V \)
Now, we differentiate \( U \) and \( V \) with respect to \( x \):
- For \( U \):
\[ \frac{dU}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}(x)) = 2 \cdot \frac{1}{1+x^2} \]
Thus:
\[ \frac{dU}{dx} = \frac{2}{1+x^2} \]

- For \( V \):
\[ \frac{dV}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}(x)) = 2 \cdot \frac{1}{1+x^2} \]
Thus:
\[ \frac{dV}{dx} = \frac{2}{1+x^2} \]

### Step 5: Find \( \frac{dU}{dV} \)
Now, we can find \( \frac{dU}{dV} \):
\[ \frac{dU}{dV} = \frac{\frac{dU}{dx}}{\frac{dV}{dx}} = \frac{\frac{2}{1+x^2}}{\frac{2}{1+x^2}} = 1 \]

### Final Result
Thus, we conclude that:
\[ \frac{d}{dV}\left(\tan^{-1}\left(\frac{2x}{1-x^2}\right)\right) = 1 \]