Differentiate `tan^(-1)((2x)/(1-x^2))` with respect to `sin^(-1)((2x)/(1+x^2))` , if `x in (-1,\ 1)`

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: Simplify the Functions
Using trigonometric identities, we can express \( u \) and \( v \) in terms of \( \tan^{-1}(x) \):
- The expression \( \frac{2x}{1-x^2} \) can be recognized as \( \tan(2\theta) \) where \( \theta = \tan^{-1}(x) \). Thus, we can write:
\[ u = \tan^{-1}\left(\frac{2\tan(\theta)}{1 - \tan^2(\theta)}\right) = 2\theta = 2\tan^{-1}(x) \]

- Similarly, for \( v \), using the identity for sine:
\[ \frac{2\tan(\theta)}{1+\tan^2(\theta)} = \sin(2\theta) \]
Thus, we can write:
\[ v = \sin^{-1}\left(\frac{2\tan(\theta)}{1+\tan^2(\theta)}\right) = 2\theta = 2\tan^{-1}(x) \]

### Step 3: Differentiate \( u \) and \( v \)
Now we differentiate \( u \) and \( v \) with respect to \( x \):
- For \( u \):
\[ \frac{du}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2} \]

- For \( v \):
\[ \frac{dv}{dx} = 2 \cdot \frac{1}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}} \cdot \frac{d}{dx}\left(\frac{2x}{1+x^2}\right) \]
To find \( \frac{d}{dx}\left(\frac{2x}{1+x^2}\right) \):
\[ \frac{d}{dx}\left(\frac{2x}{1+x^2}\right) = \frac{(1+x^2)(2) - 2x(2x)}{(1+x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1+x^2)^2} = \frac{2 - 2x^2}{(1+x^2)^2} \]
Thus,
\[ \frac{dv}{dx} = 2 \cdot \frac{1}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}} \cdot \frac{2(1-x^2)}{(1+x^2)^2} \]

### Step 4: Calculate \( \frac{du}{dv} \)
Now, we need to find \( \frac{du}{dv} \):
\[ \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{\frac{2}{1+x^2}}{\frac{dv}{dx}} \]
Since both \( u \) and \( v \) simplify to \( 2\tan^{-1}(x) \), we can conclude:
\[ \frac{du}{dv} = 1 \]

### Final Result
Thus, the derivative of \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) with respect to \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) is:
\[ \frac{du}{dv} = 1 \]