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 follow these steps:

### Step 1: Expand the function \( y \)

We start with the expression:
\[ y = (1+x)(1+x^2)(1+x^4) \]

First, we can multiply the first two factors:
\[ (1+x)(1+x^2) = 1 + x + x^2 + x^3 \]

Now, we multiply this result by the third factor \( (1+x^4) \):
\[ y = (1 + x + x^2 + x^3)(1 + x^4) \]

Expanding this, we get:
\[ y = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 \]

### Step 2: Differentiate \( y \)

Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 0 + 1 + 2x + 3x^2 + 4x^3 + 5x^4 + 6x^5 + 7x^6 \]
Thus,
\[ \frac{dy}{dx} = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + 6x^5 + 7x^6 \]

### Step 3: Evaluate \( \frac{dy}{dx} \) at \( x = 1 \)

Now we substitute \( x = 1 \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{x=1} = 1 + 2(1) + 3(1^2) + 4(1^3) + 5(1^4) + 6(1^5) + 7(1^6) \]

Calculating this gives:
\[ = 1 + 2 + 3 + 4 + 5 + 6 + 7 \]
\[ = 28 \]

### Final Answer

Thus, the value of \( \frac{dy}{dx} \) at \( x = 1 \) is:
\[ \boxed{28} \]