`y=sqrt((x^(2)+x+1)/(x^(2)-x+1))`

Answer

To find the derivative \( \frac{dy}{dx} \) for the function

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

we can follow these steps:

### Step 1: Rewrite the function
We can rewrite the function as:

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

### Step 2: Apply the logarithmic differentiation
Taking the natural logarithm of both sides gives:

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

### Step 3: Differentiate both sides
Differentiating both sides with respect to \( x \):

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

### Step 4: Multiply by \( y \)
Now, multiply both sides by \( y \):

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

### Step 5: Substitute \( y \)
Substituting back \( y = \sqrt{\frac{x^2 + x + 1}{x^2 - x + 1}} \):

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

### Step 6: Simplify the expression
Now, we simplify the expression inside the parentheses:

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

### Step 7: Final expression
After simplification, we can express \( \frac{dy}{dx} \) as:

\[ \frac{dy}{dx} = \sqrt{\frac{x^2 + x + 1}{x^2 - x + 1}} \cdot \frac{1}{2} \cdot \frac{(2x^3 + 2x^2 + 2x - 2x^2 - 2x + 1) - (2x^3 + 2x^2 + 2x - x^2 - x - 1)}{(x^2 + x + 1)(x^2 - x + 1)}. \]

This leads to the final expression for \( \frac{dy}{dx} \).