Question

16. If f(x) = (1 + x) (1 + x²) (1 + x⁴) (1 + x²), then f'(1) =

• A. 60
• B. 240
• C. 80
• D. 120

Answer 1

Answer: (D)

120

Answer 2

We interpret the intended product as

$$f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)$$

which is a standard factorization giving

$$f(x)=1+x+x^2+\cdots+x^{15}\,.$$

Step 1: Note that at $x=1$ we have

$$f(1)= (2)(2)(2)(2)=16\,.$$

Step 2: Logarithmic differentiation
Take logarithm:

$$\ln(f(x))=\ln(1+x)+\ln(1+x^2)+\ln(1+x^4)+\ln(1+x^8)$$

Differentiate 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 3: Evaluate at $x=1$

$$\frac{f'(1)}{16} = \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} = 0.5 + 1 + 2 + 4 = 7.5\,.$$

Thus,

$$f'(1)= 16\times 7.5=120\,.$$