Find the derivative of `tan^(-1) "" (sqrt(1 + x^(2)) - 1)/( x)` with respect to `tan^(-1) ( 2 x sqrt( 1 - x^(2)))/(1 - 2 x ^(2)) ` at x = 0

Answer

To find the derivative of the function

\[ y = \tan^{-1}\left(\frac{\sqrt{1 + x^2} - 1}{x}\right) \]

with respect to

\[ z = \tan^{-1}\left(\frac{2x\sqrt{1 - x^2}}{1 - 2x^2}\right) \]

at \( x = 0 \), we will follow these steps:

### Step 1: Differentiate \( y \) with respect to \( x \)

Let

\[ u = \frac{\sqrt{1 + x^2} - 1}{x} \]

Then

\[ y = \tan^{-1}(u) \]

Using the chain rule, we have:

\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]

Now we need to find \( \frac{du}{dx} \).

### Step 2: Differentiate \( u \)

To differentiate \( u \):

\[ u = \frac{\sqrt{1 + x^2} - 1}{x} \]

Using the quotient rule:

\[ \frac{du}{dx} = \frac{x \cdot \frac{d}{dx}(\sqrt{1 + x^2} - 1) - (\sqrt{1 + x^2} - 1) \cdot \frac{d}{dx}(x)}{x^2} \]

Calculating \( \frac{d}{dx}(\sqrt{1 + x^2}) \):

\[ \frac{d}{dx}(\sqrt{1 + x^2}) = \frac{x}{\sqrt{1 + x^2}} \]

Thus,

\[ \frac{du}{dx} = \frac{x \cdot \frac{x}{\sqrt{1 + x^2}} - (\sqrt{1 + x^2} - 1)}{x^2} \]

Simplifying:

\[ \frac{du}{dx} = \frac{x^2/\sqrt{1 + x^2} - \sqrt{1 + x^2} + 1}{x^2} \]

### Step 3: Evaluate \( \frac{du}{dx} \) at \( x = 0 \)

At \( x = 0 \):

\[ u = \frac{\sqrt{1 + 0^2} - 1}{0} \text{ (indeterminate form)} \]

Using L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{\sqrt{1 + x^2} - 1}{x} = \lim_{x \to 0} \frac{\frac{x}{\sqrt{1 + x^2}}}{1} = 0 \]

So, \( u(0) = 0 \).

Now, substituting \( x = 0 \) into \( \frac{du}{dx} \):

\[ \frac{du}{dx} \text{ at } x = 0 = \frac{0 - (1 - 1)}{0} = 0 \]

### Step 4: Differentiate \( z \) with respect to \( x \)

Let

\[ v = \frac{2x\sqrt{1 - x^2}}{1 - 2x^2} \]

Then

\[ z = \tan^{-1}(v) \]

Using the chain rule:

\[ \frac{dz}{dx} = \frac{1}{1 + v^2} \cdot \frac{dv}{dx} \]

Now we need to find \( \frac{dv}{dx} \).

### Step 5: Differentiate \( v \)

Using the quotient rule:

\[ \frac{dv}{dx} = \frac{(1 - 2x^2)(2\sqrt{1 - x^2} + 2x \cdot \frac{-x}{\sqrt{1 - x^2}}) - 2x\sqrt{1 - x^2}(-4x)}{(1 - 2x^2)^2} \]

### Step 6: Evaluate \( \frac{dv}{dx} \) at \( x = 0 \)

At \( x = 0 \):

\[ v = 0 \quad \text{and} \quad \frac{dv}{dx} \text{ at } x = 0 = 2 \]

### Step 7: Combine results

Now we can find the derivative:

\[ \frac{dy}{dz} = \frac{\frac{dy}{dx}}{\frac{dz}{dx}} = \frac{\frac{1}{1 + u^2} \cdot \frac{du}{dx}}{\frac{1}{1 + v^2} \cdot \frac{dv}{dx}} \]

At \( x = 0 \):

\[ \frac{dy}{dx} = \frac{1}{1 + 0^2} \cdot 0 = 0 \]

\[ \frac{dz}{dx} = \frac{1}{1 + 0^2} \cdot 2 = 2 \]

Thus,

\[ \frac{dy}{dz} = \frac{0}{2} = 0 \]

### Final Answer

The derivative at \( x = 0 \) is

\[ \frac{dy}{dz} = 0 \]