Given that,
`(d)/(dx)((1+x^(2)+x^(4))/(1+x+x^(2)))=Ax+B`
What is the value of B ?

Options

A. `-1`
B. 1
C. 2
D. 4

Answer

To solve the problem, we need to find the derivative of the function \((1 + x^2 + x^4)/(1 + x + x^2)\) and express it in the form \(Ax + B\). We will then determine the value of \(B\).

### Step-by-Step Solution:

1. **Identify the Function**:
We have the function:
\[ f(x) = \frac{1 + x^2 + x^4}{1 + x + x^2} \]

2. **Differentiate Using the Quotient Rule**:
The quotient rule states that if \(f(x) = \frac{g(x)}{h(x)}\), then:
\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \]
Here, \(g(x) = 1 + x^2 + x^4\) and \(h(x) = 1 + x + x^2\).

3. **Calculate \(g'(x)\) and \(h'(x)\)**:
- For \(g(x)\):
\[ g'(x) = 0 + 2x + 4x^3 = 2x + 4x^3 \]
- For \(h(x)\):
\[ h'(x) = 0 + 1 + 2x = 1 + 2x \]

4. **Substitute into the Quotient Rule**:
Now substituting \(g(x)\), \(g'(x)\), \(h(x)\), and \(h'(x)\) into the quotient rule:
\[ f'(x) = \frac{(2x + 4x^3)(1 + x + x^2) - (1 + x^2 + x^4)(1 + 2x)}{(1 + x + x^2)^2} \]

5. **Simplify the Numerator**:
We need to expand and simplify the numerator:
- Expand \((2x + 4x^3)(1 + x + x^2)\):
\[ = 2x + 2x^2 + 2x^3 + 4x^3 + 4x^4 + 4x^5 = 2x + 2x^2 + 6x^3 + 4x^4 + 4x^5 \]
- Expand \((1 + x^2 + x^4)(1 + 2x)\):
\[ = 1 + 2x + x^2 + 2x^3 + x^4 + 2x^5 = 1 + 2x + x^2 + 2x^3 + x^4 + 2x^5 \]
- Now, subtract the second expansion from the first:
\[ (2x + 2x^2 + 6x^3 + 4x^4 + 4x^5) - (1 + 2x + x^2 + 2x^3 + x^4 + 2x^5) \]
This simplifies to:
\[ (2x - 2x) + (2x^2 - x^2) + (6x^3 - 2x^3) + (4x^4 - x^4) + (4x^5 - 2x^5) - 1 \]
Which results in:
\[ 0 + x^2 + 4x^3 + 3x^4 + 2x^5 - 1 \]

6. **Final Form**:
The derivative simplifies to:
\[ f'(x) = \frac{x^2 + 4x^3 + 3x^4 + 2x^5 - 1}{(1 + x + x^2)^2} \]
We need to express this in the form \(Ax + B\).

7. **Compare Coefficients**:
From the previous steps, we can see that after simplification, the linear term is \(2x\) and the constant term is \(-1\). Therefore, we have:
\[ A = 2 \quad \text{and} \quad B = -1 \]

### Conclusion:
The value of \(B\) is:
\[ \boxed{-1} \]