If `y=(x^(2)+x+1)/(x^(2)-x+1)," then "(x^(2)-x+1)^(2)y_(1)=`

Options

A. `2(1-x^(2))`
B. `1-x^(2)`
C. `x^(2)-1`
D. `2(x-1)`

Answer

To solve the problem, we need to differentiate the function \( y = \frac{x^2 + x + 1}{x^2 - x + 1} \) and then multiply the derivative by \( (x^2 - x + 1)^2 \).

### Step-by-Step Solution:

1. **Identify the functions**:
Let \( u = x^2 + x + 1 \) (the numerator) and \( v = x^2 - x + 1 \) (the denominator).

2. **Differentiate using the Quotient Rule**:
The Quotient Rule states that if \( y = \frac{u}{v} \), then:
\[ y' = \frac{v \cdot u' - u \cdot v'}{v^2} \]
where \( u' \) is the derivative of \( u \) and \( v' \) is the derivative of \( v \).

3. **Calculate \( u' \) and \( v' \)**:
\[ u' = \frac{d}{dx}(x^2 + x + 1) = 2x + 1 \]
\[ v' = \frac{d}{dx}(x^2 - x + 1) = 2x - 1 \]

4. **Substitute into the Quotient Rule**:
Now substitute \( u \), \( u' \), \( v \), and \( v' \) into the Quotient Rule formula:
\[ y' = \frac{(x^2 - x + 1)(2x + 1) - (x^2 + x + 1)(2x - 1)}{(x^2 - x + 1)^2} \]

5. **Expand the numerator**:
Expanding both terms in the numerator:
- First term:
\[ (x^2 - x + 1)(2x + 1) = 2x^3 + x^2 - 2x^2 - x + 2x + 1 = 2x^3 - x^2 + x + 1 \]
- Second term:
\[ (x^2 + x + 1)(2x - 1) = 2x^3 + 2x^2 + 2x - x^2 - x - 1 = 2x^3 + x^2 + x - 1 \]

6. **Combine the terms**:
Now combine the two expanded terms:
\[ y' = \frac{(2x^3 - x^2 + x + 1) - (2x^3 + x^2 + x - 1)}{(x^2 - x + 1)^2} \]
Simplifying the numerator:
\[ = \frac{2}{(x^2 - x + 1)^2} \]

7. **Multiply by \( (x^2 - x + 1)^2 \)**:
Now, we need to find \( (x^2 - x + 1)^2 y' \):
\[ (x^2 - x + 1)^2 y' = (x^2 - x + 1)^2 \cdot \frac{2}{(x^2 - x + 1)^2} = 2 \]

### Final Answer:
\[ (x^2 - x + 1)^2 y' = 2 \]