If `y=(1+x)(1+x^2)(1+x^4)`, then` (dy)/(dx)` at x =1 is :

Options

A. 20
B. 28
C. 1
D. 0

Answer

To find \(\frac{dy}{dx}\) at \(x = 1\) for the function \(y = (1+x)(1+x^2)(1+x^4)\), we will use the product rule for differentiation.

### Step-by-Step Solution

1. **Identify the function**:
\[ y = (1+x)(1+x^2)(1+x^4) \]

2. **Apply the product rule**:
The product rule states that if \(y = u \cdot v \cdot w\), then:
\[ \frac{dy}{dx} = u'vw + uv'w + uvw' \]
where \(u = (1+x)\), \(v = (1+x^2)\), and \(w = (1+x^4)\).

3. **Differentiate each function**:
- \(u' = \frac{d}{dx}(1+x) = 1\)
- \(v' = \frac{d}{dx}(1+x^2) = 2x\)
- \(w' = \frac{d}{dx}(1+x^4) = 4x^3\)

4. **Substitute into the product rule**:
\[ \frac{dy}{dx} = (1)(1+x^2)(1+x^4) + (1+x)(2x)(1+x^4) + (1+x)(1+x^2)(4x^3) \]

5. **Simplify each term**:
- First term:
\[ (1)(1+x^2)(1+x^4) = (1+x^2)(1+x^4) = 1 + x^4 + x^2 + x^6 \]
- Second term:
\[ (1+x)(2x)(1+x^4) = 2x(1+x)(1+x^4) = 2x(1 + x + x^4 + x^5) = 2x + 2x^2 + 2x^5 + 2x^6 \]
- Third term:
\[ (1+x)(1+x^2)(4x^3) = 4x^3(1+x)(1+x^2) = 4x^3(1 + x + x^2 + x^3) = 4x^3 + 4x^4 + 4x^5 + 4x^6 \]

6. **Combine all terms**:
\[ \frac{dy}{dx} = (1 + x^4 + x^2 + x^6) + (2x + 2x^2 + 2x^5 + 2x^6) + (4x^3 + 4x^4 + 4x^5 + 4x^6) \]
Combine like terms:
\[ = 1 + 2x + 5x^2 + 8x^3 + 7x^4 + 6x^5 + 5x^6 \]

7. **Evaluate at \(x = 1\)**:
\[ \frac{dy}{dx} \bigg|_{x=1} = 1 + 2(1) + 5(1) + 8(1) + 7(1) + 6(1) + 5(1) = 1 + 2 + 5 + 8 + 7 + 6 + 5 = 34 \]

### Final Answer:
\[ \frac{dy}{dx} \text{ at } x = 1 \text{ is } 34. \]