`cos^(-1)((x-x^1)/(x+x^(-1)))`

Answer

To differentiate the function \( y = \cos^{-1} \left( \frac{x - \frac{1}{x}}{x + \frac{1}{x}} \right) \), we will follow these steps:

### Step 1: Simplify the Argument of the Inverse Cosine
First, we simplify the expression inside the inverse cosine function:

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

Thus, we can rewrite the function as:

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

### Step 2: Differentiate Using the Chain Rule
Now, we differentiate \( y \) with respect to \( x \). Using the chain rule, we have:

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

### Step 3: Differentiate the Inner Function
Next, we differentiate the inner function \( \frac{x^2 - 1}{x^2 + 1} \) using the quotient rule:

Let \( u = x^2 - 1 \) and \( v = x^2 + 1 \). Then:

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

Calculating \( \frac{du}{dx} = 2x \) and \( \frac{dv}{dx} = 2x \):

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

Simplifying the numerator:

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

### Step 4: Substitute Back into the Derivative
Now substituting back into the derivative:

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

### Step 5: Simplify the Square Root
Next, we simplify \( 1 - \left( \frac{x^2 - 1}{x^2 + 1} \right)^2 \):

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

Calculating the numerator:

\[ = (x^2 + 1 + x^2 - 1)(x^2 + 1 - (x^2 - 1)) = (2x^2)(2) = 4x^2 \]

Thus:

\[ 1 - \left( \frac{x^2 - 1}{x^2 + 1} \right)^2 = \frac{4x^2}{(x^2 + 1)^2} \]

### Step 6: Final Derivative Expression
Now substituting this back into the derivative:

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

This simplifies to:

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

### Final Answer
Thus, the derivative of the given function is:

\[ \frac{dy}{dx} = -\frac{2}{x^2 + 1} \]