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

Options

A. `(dy)/(dx)=2x^(2)`
B. `(dy)/(dx)=3x^(2)`
C. `(dy)/(dx)=5x^(2)`
D. `(dy)/(dx)=3x`

Answer

To find the derivative of the function \( y = (x - 1)(x^2 + x + 1) \), we can use the product rule of differentiation. The product rule states that if you have two functions \( f(x) \) and \( g(x) \), then the derivative of their product is given by:

\[ \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) \]

### Step-by-Step Solution:

1. **Identify the functions**:
Let \( f(x) = x - 1 \) and \( g(x) = x^2 + x + 1 \).

2. **Find the derivatives**:
- The derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x - 1) = 1 \]
- The derivative of \( g(x) \):
\[ g'(x) = \frac{d}{dx}(x^2 + x + 1) = 2x + 1 \]

3. **Apply the product rule**:
Using the product rule:
\[ y' = f'(x)g(x) + f(x)g'(x) \]
Substitute \( f(x) \), \( g(x) \), \( f'(x) \), and \( g'(x) \):
\[ y' = (1)(x^2 + x + 1) + (x - 1)(2x + 1) \]

4. **Simplify the expression**:
- First part:
\[ y' = x^2 + x + 1 \]
- Second part:
\[ (x - 1)(2x + 1) = 2x^2 + x - 2x - 1 = 2x^2 - x - 1 \]
- Combine both parts:
\[ y' = (x^2 + x + 1) + (2x^2 - x - 1) \]
\[ y' = x^2 + 2x^2 + x - x + 1 - 1 \]
\[ y' = 3x^2 \]

5. **Final result**:
The derivative of the function \( y \) is:
\[ y' = 3x^2 \]