Differentiate `tan^(-1)((sqrt(1+x^(2))+1)/(x))" w.r.t. "tan^(-1)((2xsqrt(1-x^(2)))/(1-2x^(2)))` at x = 0.

Answer

To differentiate \( u = \tan^{-1}\left(\frac{\sqrt{1+x^2}+1}{x}\right) \) with respect to \( v = \tan^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right) \) at \( x = 0 \), we will use the chain rule and implicit differentiation.

### Step-by-Step Solution:

1. **Define the Functions**:
Let:
\[ u = \tan^{-1}\left(\frac{\sqrt{1+x^2}+1}{x}\right) \]
\[ v = \tan^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right) \]

2. **Differentiate \( u \) with respect to \( x \)**:
Using the derivative of the inverse tangent function:
\[ \frac{du}{dx} = \frac{1}{1 + \left(\frac{\sqrt{1+x^2}+1}{x}\right)^2} \cdot \frac{d}{dx}\left(\frac{\sqrt{1+x^2}+1}{x}\right) \]
Now, we need to differentiate \( \frac{\sqrt{1+x^2}+1}{x} \):
\[ \frac{d}{dx}\left(\frac{\sqrt{1+x^2}+1}{x}\right) = \frac{x \cdot \frac{d}{dx}(\sqrt{1+x^2}+1) - (\sqrt{1+x^2}+1) \cdot \frac{d}{dx}(x)}{x^2} \]
\[ = \frac{x \cdot \frac{x}{\sqrt{1+x^2}} - (\sqrt{1+x^2}+1)}{x^2} \]

3. **Evaluate \( \frac{du}{dx} \) at \( x = 0 \)**:
At \( x = 0 \):
\[ u = \tan^{-1}\left(\frac{\sqrt{1+0^2}+1}{0}\right) \text{ (undefined, but we can find the limit)} \]
As \( x \to 0 \), \( u \to \tan^{-1}(\infty) = \frac{\pi}{2} \).

4. **Differentiate \( v \) with respect to \( x \)**:
Similarly, for \( v \):
\[ \frac{dv}{dx} = \frac{1}{1 + \left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)^2} \cdot \frac{d}{dx}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right) \]
Differentiate \( \frac{2x\sqrt{1-x^2}}{1-2x^2} \) using the quotient rule.

5. **Evaluate \( \frac{dv}{dx} \) at \( x = 0 \)**:
At \( x = 0 \):
\[ v = \tan^{-1}(0) = 0 \]

6. **Find \( \frac{du}{dv} \)**:
Using the chain rule:
\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} \]
Substitute the values of \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) evaluated at \( x = 0 \).

7. **Final Calculation**:
Calculate \( \frac{du}{dv} \) at \( x = 0 \) to get the final result.