If ` f(x) = (1 + x) (1 + x^(2)) (1 + x^(4)) (1 + x^(8)) ` , then f'(1) is

Options

A. 130
B. 120
C. 110
D. None of these

Explanation

To find \( f'(1) \) for the function \( f(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8) \), we will follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8) \]

### Step 2: Take the natural logarithm
Taking the natural logarithm of both sides gives:
\[ \log f(x) = \log(1 + x) + \log(1 + x^2) + \log(1 + x^4) + \log(1 + x^8) \]

### Step 3: Differentiate both sides
Now we differentiate both sides with respect to \( x \):
\[ \frac{f'(x)}{f(x)} = \frac{1}{1 + x} + \frac{2x}{1 + x^2} + \frac{4x^3}{1 + x^4} + \frac{8x^7}{1 + x^8} \]

### Step 4: Solve for \( f'(x) \)
Multiplying both sides by \( f(x) \) gives:
\[ f'(x) = f(x) \left( \frac{1}{1 + x} + \frac{2x}{1 + x^2} + \frac{4x^3}{1 + x^4} + \frac{8x^7}{1 + x^8} \right) \]

### Step 5: Evaluate \( f(1) \)
Next, we calculate \( f(1) \):
\[ f(1) = (1 + 1)(1 + 1^2)(1 + 1^4)(1 + 1^8) = 2 \cdot 2 \cdot 2 \cdot 2 = 16 \]

### Step 6: Evaluate \( f'(1) \)
Now we substitute \( x = 1 \) into the derivative:
\[ f'(1) = f(1) \left( \frac{1}{1 + 1} + \frac{2 \cdot 1}{1 + 1^2} + \frac{4 \cdot 1^3}{1 + 1^4} + \frac{8 \cdot 1^7}{1 + 1^8} \right) \]
Calculating each term:
- \( \frac{1}{2} \)
- \( \frac{2}{2} = 1 \)
- \( \frac{4}{2} = 2 \)
- \( \frac{8}{2} = 4 \)

Adding these gives:
\[ \frac{1}{2} + 1 + 2 + 4 = \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} = \frac{15}{2} \]

### Step 7: Final calculation of \( f'(1) \)
Thus, we have:
\[ f'(1) = 16 \cdot \frac{15}{2} = 120 \]

### Conclusion
The value of \( f'(1) \) is:
\[ \boxed{120} \]