Differentiate `tan^(-1)'(sqrt(1+x^(2))-1)/(x)` w.r.t. `tan^(-1)x`, when `x ne 0`.

Explanation

To differentiate the expression \( \frac{\tan^{-1}(\sqrt{1+x^2}-1)}{x} \) with respect to \( \tan^{-1}(x) \) when \( x \neq 0 \), we will follow these steps:

### Step 1: Define the functions
Let:
- \( u = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \)
- \( v = \tan^{-1}(x) \)

### Step 2: Simplify the expression for \( u \)
We start by simplifying \( u \):
1. Recognize that \( \sqrt{1+x^2} = \sec(\theta) \) when \( x = \tan(\theta) \).
2. Thus, we can rewrite \( u \):
\[ u = \tan^{-1}\left(\frac{\sec(\theta)-1}{\tan(\theta)}\right) \]
3. Using the identity \( \sec^2(\theta) = 1 + \tan^2(\theta) \), we can express \( \sec(\theta) \) in terms of \( \tan(\theta) \).
4. This gives:
\[ u = \tan^{-1}\left(\frac{\sec(\theta)-1}{\tan(\theta)}\right) = \tan^{-1}\left(\frac{\sqrt{1+\tan^2(\theta)}-1}{\tan(\theta)}\right) \]

### Step 3: Further simplification
1. We can express \( \sec(\theta) \) as \( \frac{1}{\cos(\theta)} \) and \( \tan(\theta) \) as \( \frac{\sin(\theta)}{\cos(\theta)} \).
2. This leads to:
\[ u = \tan^{-1}\left(\frac{\frac{1}{\cos(\theta)} - 1}{\frac{\sin(\theta)}{\cos(\theta)}}\right) = \tan^{-1}\left(\frac{1 - \cos(\theta)}{\sin(\theta)}\right) \]
3. Using the identity \( 1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right) \), we can write:
\[ u = \tan^{-1}\left(\frac{2\sin^2\left(\frac{\theta}{2}\right)}{\sin(\theta)}\right) \]
4. Since \( \sin(\theta) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \), we find:
\[ u = \tan^{-1}\left(\frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)}\right) = \frac{\theta}{2} \]
5. Substituting back \( \theta = \tan^{-1}(x) \):
\[ u = \frac{1}{2}\tan^{-1}(x) \]

### Step 4: Differentiate \( u \) and \( v \)
Now we differentiate \( u \) and \( v \):
1. Differentiate \( u \):
\[ \frac{du}{dx} = \frac{1}{2} \cdot \frac{1}{1+x^2} \]
2. Differentiate \( v \):
\[ \frac{dv}{dx} = \frac{1}{1+x^2} \]

### Step 5: Find \( \frac{du}{dv} \)
Using the chain rule:
\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{1}{2(1+x^2)}}{\frac{1}{1+x^2}} = \frac{1}{2} \]

### Final Answer
Thus, the derivative \( \frac{du}{dv} \) is:
\[ \frac{1}{2} \]