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 will follow these steps:

### Step 1: Substitute \( x = \tan(\theta) \)

Let \( x = \tan(\theta) \). Then we can express \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) and \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) in terms of \( \theta \).

### Step 2: Simplify \( \tan^{-1}\left(\frac{2\tan(\theta)}{1-\tan^2(\theta)}\right) \)

Using the double angle formula for tangent, we have:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1-\tan^2(\theta)} \]
Thus,
\[ \tan^{-1}\left(\frac{2\tan(\theta)}{1-\tan^2(\theta)}\right) = 2\theta \]

### Step 3: Simplify \( \sin^{-1}\left(\frac{2\tan(\theta)}{1+\tan^2(\theta)}\right) \)

Using the double angle formula for sine, we have:
\[ \sin(2\theta) = \frac{2\tan(\theta)}{1+\tan^2(\theta)} \]
Thus,
\[ \sin^{-1}\left(\frac{2\tan(\theta)}{1+\tan^2(\theta)}\right) = 2\theta \]

### Step 4: Differentiate \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) and \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \)

Now we differentiate both expressions with respect to \( \theta \):
1. The derivative of \( 2\theta \) with respect to \( \theta \) is \( 2 \).
2. The derivative of \( 2\theta \) with respect to \( \theta \) is also \( 2 \).

### Step 5: Apply the chain rule

Using the chain rule, we find:
\[ \frac{d}{d\theta}\left(\tan^{-1}\left(\frac{2x}{1-x^2}\right)\right) = 2 \]
\[ \frac{d}{d\theta}\left(\sin^{-1}\left(\frac{2x}{1+x^2}\right)\right) = 2 \]

### Step 6: Find the derivative of \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) with respect to \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \)

Thus, we have:
\[ \frac{d}{d\theta}\left(\tan^{-1}\left(\frac{2x}{1-x^2}\right)\right) \div \frac{d}{d\theta}\left(\sin^{-1}\left(\frac{2x}{1+x^2}\right)\right) = \frac{2}{2} = 1 \]

### Final Answer

Therefore, 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 \( 1 \).

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