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

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 start by rewriting the function in a more manageable form:
\[ y = \left( \frac{x^2 + x + 1}{x^2 - x + 1} \right)^{\frac{1}{2}} \]

### Step 2: Take the logarithm of both sides
Taking the natural logarithm of both sides gives us:
\[ \log y = \frac{1}{2} \log \left( \frac{x^2 + x + 1}{x^2 - x + 1} \right) \]

### Step 3: Apply logarithmic properties
Using the properties of logarithms, we can expand the right-hand side:
\[ \log y = \frac{1}{2} \left( \log(x^2 + x + 1) - \log(x^2 - x + 1) \right) \]

### Step 4: Differentiate both sides
Now, we differentiate both sides with respect to \( x \):
\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left( \frac{1}{x^2 + x + 1} \cdot (2x + 1) - \frac{1}{x^2 - x + 1} \cdot (2x - 1) \right) \]

### Step 5: Simplify the right-hand side
This expression simplifies to:
\[ \frac{dy}{dx} = y \cdot \frac{1}{2} \left( \frac{2x + 1}{x^2 + x + 1} - \frac{2x - 1}{x^2 - x + 1} \right) \]

### Step 6: Find a common denominator
To combine the fractions, we find a common denominator:
\[ \frac{dy}{dx} = y \cdot \frac{1}{2} \cdot \frac{(2x + 1)(x^2 - x + 1) - (2x - 1)(x^2 + x + 1)}{(x^2 + x + 1)(x^2 - x + 1)} \]

### Step 7: Expand and simplify the numerator
Expanding the numerator:
\[ (2x + 1)(x^2 - x + 1) = 2x^3 - 2x^2 + 2x + x^2 - x + 1 = 2x^3 - x^2 + x + 1 \]
\[ (2x - 1)(x^2 + x + 1) = 2x^3 + 2x^2 + 2x - x^2 - x - 1 = 2x^3 + x^2 + x - 1 \]

Subtracting these gives:
\[ (2x^3 - x^2 + x + 1) - (2x^3 + x^2 + x - 1) = -2x^2 + 2 \]

### Step 8: Substitute back for \( y \)
Now substituting back for \( y \):
\[ \frac{dy}{dx} = \sqrt{\frac{x^2 + x + 1}{x^2 - x + 1}} \cdot \frac{-2x^2 + 2}{2(x^2 + x + 1)(x^2 - x + 1)} \]

### Final Result
Thus, the derivative is:
\[ \frac{dy}{dx} = \sqrt{\frac{x^2 + x + 1}{x^2 - x + 1}} \cdot \frac{1 - x^2}{x^4 + x^2 + 1} \]