If ` (d)/(dx) ((1 +x^2 +x^4)/( 1+x+x^2))=ax +b ` then ` a= ` and b=…..

Answer

To solve the problem, we need to differentiate the function given by:

\[ f(x) = \frac{1 + x^2 + x^4}{1 + x + x^2} \]

and set the result equal to \( ax + b \). We will find the values of \( a \) and \( b \).

### Step 1: Differentiate the function using the Quotient Rule

The Quotient Rule states that if you have a function \( \frac{u}{v} \), then its derivative is given by:

\[ \frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Here, let:
- \( u = 1 + x^2 + x^4 \)
- \( v = 1 + x + x^2 \)

Now we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

### Step 2: Calculate \( \frac{du}{dx} \)

\[ u = 1 + x^2 + x^4 \]
Differentiating \( u \):

\[ \frac{du}{dx} = 0 + 2x + 4x^3 = 2x + 4x^3 \]

### Step 3: Calculate \( \frac{dv}{dx} \)

\[ v = 1 + x + x^2 \]
Differentiating \( v \):

\[ \frac{dv}{dx} = 0 + 1 + 2x = 1 + 2x \]

### Step 4: Apply the Quotient Rule

Now substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the Quotient Rule:

\[ \frac{d}{dx} \left(\frac{u}{v}\right) = \frac{(1 + x + x^2)(2x + 4x^3) - (1 + x^2 + x^4)(1 + 2x)}{(1 + x + x^2)^2} \]

### Step 5: Simplify the numerator

Now we will expand and simplify the numerator:

1. Expand \( (1 + x + x^2)(2x + 4x^3) \):
\[ = 2x + 4x^3 + 2x^2 + 4x^4 \]
\[ = 2x + 2x^2 + 4x^3 + 4x^4 \]

2. 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:

\[ (2x + 2x^2 + 4x^3 + 4x^4) - (1 + 2x + x^2 + 2x^3 + x^4 + 2x^5) \]

### Step 6: Combine like terms

Combining like terms gives:

- The constant term: \( -1 \)
- The \( x \) term: \( 2x - 2x = 0 \)
- The \( x^2 \) term: \( 2x^2 - x^2 = x^2 \)
- The \( x^3 \) term: \( 4x^3 - 2x^3 = 2x^3 \)
- The \( x^4 \) term: \( 4x^4 - x^4 = 3x^4 \)
- The \( x^5 \) term: \( -2x^5 \)

So the numerator simplifies to:

\[ x^2 + 2x^3 + 3x^4 - 1 - 2x^5 \]

### Step 7: Set the result equal to \( ax + b \)

Now we need to compare the simplified derivative with \( ax + b \).

From the simplified expression, we can see that the linear term is missing, hence:

- \( a = 0 \) (coefficient of \( x \))
- \( b = -1 \) (constant term)

### Final Result

Thus, we have:

\[ a = 0, \quad b = -1 \]