What is $f'(1)$ for the function $f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)$?

24 60 48 120

120

$f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)$. Taking logarithm in both sides we get $\log(f(x)=\log(1+x)+\log(1+x^2)+\log(1+x^4)+\log(1+x^8)$. Differentiating both sides with respect to $x$ we get $\frac{f'(x)}{f(x)}=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}$. Now substituting $x=1$ we get the answer