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

Explanation

To differentiate the function \( y = \tan^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} \right) \) with respect to \( z = \sin^{-1} \left( \frac{2x}{1+x^2} \right) \), we can use the chain rule. Let's break down the steps:

### Step 1: Rewrite \( x \) in terms of \( \theta \)
Let \( x = \tan(\theta) \). Then, we can express \( \sin(2\theta) \) as follows:
\[ \sin(2\theta) = \frac{2\tan(\theta)}{1+\tan^2(\theta)} = \frac{2x}{1+x^2} \]
This means \( z = \sin^{-1} \left( \frac{2x}{1+x^2} \right) = 2\theta \).
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