Differentiate `tan^(-1)((sqrt(1+x^2)-1)/x)w.r.t tan^(-1)x ,w h e r ex!=0.`

Explanation

To differentiate \( u = \tan^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} \right) \) with respect to \( v = \tan^{-1}(x) \), we will follow these steps:

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

We need to find \( \frac{du}{dv} \).

### Step 2: Use the chain rule
Using the chain rule, we have:
\[ \frac{du}{dv} = \frac{du}{dx} \cdot \frac{dx}{dv} \]
Since \( v = \tan^{-1}(x) \), we can find \( \frac{dx}{dv} \) as follows:
\[ \frac{dx}{dv} = \frac{1}{\frac{dv}{dx}} = \frac{1}{\frac{1}{1+x^2}} = 1+x^2 \]

### Step 3: Differentiate \( u \) with respect to \( x \)
To find \( \frac{du}{dx} \), we first need to differentiate \( u \):
\[ u = \tan^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} \right) \]
Using the derivative of the inverse tangent function:
\[ \frac{du}{dx} = \frac{1}{1 + \left( \frac{\sqrt{1+x^2}-1}{x} \right)^2} \cdot \frac{d}{dx} \left( \frac{\sqrt{1+x^2}-1}{x} \right) \]

### Step 4: Differentiate the inner function
Let \( y = \frac{\sqrt{1+x^2}-1}{x} \). We need to differentiate \( y \):
Using the quotient rule:
\[ \frac{dy}{dx} = \frac{x \cdot \frac{d}{dx}(\sqrt{1+x^2}-1) - (\sqrt{1+x^2}-1) \cdot \frac{d}{dx}(x)}{x^2} \]
Calculating \( \frac{d}{dx}(\sqrt{1+x^2}-1) \):
\[ \frac{d}{dx}(\sqrt{1+x^2}) = \frac{x}{\sqrt{1+x^2}} \]
Thus:
\[ \frac{dy}{dx} = \frac{x \cdot \frac{x}{\sqrt{1+x^2}} - \left(\sqrt{1+x^2}-1\right)}{x^2} \]

### Step 5: Substitute back into \( \frac{du}{dx} \)
Now substitute \( \frac{dy}{dx} \) back into \( \frac{du}{dx} \):
\[ \frac{du}{dx} = \frac{1}{1 + \left( \frac{\sqrt{1+x^2}-1}{x} \right)^2} \cdot \frac{dy}{dx} \]

### Step 6: Combine to find \( \frac{du}{dv} \)
Now substitute \( \frac{du}{dx} \) and \( \frac{dx}{dv} \) into the chain rule:
\[ \frac{du}{dv} = \frac{du}{dx} \cdot (1+x^2) \]

### Step 7: Simplify the expression
After substituting and simplifying, we find:
\[ \frac{du}{dv} = \frac{1}{2} \]

### Final Answer
Thus, the final result is:
\[ \frac{du}{dv} = \frac{1}{2} \]