If `y=(1)/(1+x+x^(2))`, then `(dy)/(dx)=`

Options

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

Answer

To find the derivative \(\frac{dy}{dx}\) for the function \(y = \frac{1}{1 + x + x^2}\), we will use the quotient rule for differentiation. The quotient rule states that if we have a function in the form \(y = \frac{f(x)}{g(x)}\), then the derivative is given by:

\[ \frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \]

### Step-by-Step Solution:

1. **Identify \(f(x)\) and \(g(x)\)**:
- Here, \(f(x) = 1\) and \(g(x) = 1 + x + x^2\).

2. **Compute \(f'(x)\) and \(g'(x)\)**:
- \(f'(x) = 0\) (since the derivative of a constant is zero).
- \(g'(x) = 0 + 1 + 2x = 1 + 2x\).

3. **Apply the quotient rule**:
- Using the quotient rule, we have:
\[ \frac{dy}{dx} = \frac{0 \cdot g(x) - 1 \cdot g'(x)}{(g(x))^2} \]
- Substituting \(g(x)\) and \(g'(x)\):
\[ \frac{dy}{dx} = \frac{0 - (1 + 2x)}{(1 + x + x^2)^2} \]

4. **Simplify the expression**:
- This simplifies to:
\[ \frac{dy}{dx} = \frac{-(1 + 2x)}{(1 + x + x^2)^2} \]
- We can also express this in terms of \(y\):
- Since \(y = \frac{1}{1 + x + x^2}\), we can write \(1 + x + x^2 = \frac{1}{y}\).
- Therefore, \((1 + x + x^2)^2 = \frac{1}{y^2}\).

5. **Final expression**:
- Substituting this back into our derivative:
\[ \frac{dy}{dx} = -y^2(1 + 2x) \]

### Final Answer:
\[ \frac{dy}{dx} = -y^2(1 + 2x) \]