Find the derivatives of the following functions (1-3) at any point of their domains :
`y=(x-1/x)^(2)`

Answer

To find the derivative of the function \( y = \left( x - \frac{1}{x} \right)^2 \), we will use the chain rule and the power rule. Let's go through the steps:

### Step 1: Rewrite the function
We start with the function:
\[ y = \left( x - \frac{1}{x} \right)^2 \]

### Step 2: Apply the chain rule
Using the chain rule, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 2 \left( x - \frac{1}{x} \right) \cdot \frac{d}{dx} \left( x - \frac{1}{x} \right) \]

### Step 3: Differentiate the inner function
Now we need to differentiate the inner function \( x - \frac{1}{x} \):
\[ \frac{d}{dx} \left( x - \frac{1}{x} \right) = 1 + \frac{1}{x^2} \]

### Step 4: Substitute back into the derivative
Now we substitute this back into our derivative:
\[ \frac{dy}{dx} = 2 \left( x - \frac{1}{x} \right) \cdot \left( 1 + \frac{1}{x^2} \right) \]

### Step 5: Simplify the expression
Now we simplify the expression:
\[ \frac{dy}{dx} = 2 \left( x - \frac{1}{x} \right) \cdot \left( 1 + \frac{1}{x^2} \right) \]
Expanding this:
\[ = 2 \left( x - \frac{1}{x} + \frac{x}{x^2} - \frac{1}{x^3} \right) \]
\[ = 2 \left( x - \frac{1}{x} + \frac{1}{x} - \frac{1}{x^3} \right) \]
\[ = 2 \left( x - \frac{1}{x^3} \right) \]

### Final Result
Thus, the derivative of the function \( y = \left( x - \frac{1}{x} \right)^2 \) is:
\[ \frac{dy}{dx} = 2 \left( x - \frac{1}{x^3} \right) \]