Solution:

Provided: $f(x)=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)$………..(1)

$\log f(x)=\log (1+x)+\log \left(1+x^{2}\right)+\log \left(1+x^{4}\right)+\log \left(1+x^{8}\right)$

$\frac{1}{f(x)} \frac{d}{d x} f(x)=\frac{1}{1+x} \frac{d}{d x}(1+x)+\frac{1}{1+x^{2}} \frac{d}{d x}\left(1+x^{2}\right)+\frac{1}{1+x^{4}} \frac{d}{d x}\left(1+x^{4}\right)+\frac{1}{1+x^{8}} \frac{d}{d x}\left(1+x^{8}\right)$

$\frac{1}{f(x)} f^{\prime}(x)=\frac{1}{1+x} \cdot 1+\frac{1}{1+x^{2}} 2 x+\frac{1}{1+x^{4}} \cdot 4 x^{3}+\frac{1}{1+x^{8}} 8 x^{2}$

$f^{\prime}(x)=f(x)\left[\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{4 x^{3}}{1+x^{4}}+\frac{8 x^{7}}{1+x^{3}}\right]$

Putting the value of $f(x)$ from equation(1),

$f^{\prime}(x)=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)\left[\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{4 x^{3}}{1+x^{4}}+\frac{8 x^{7}}{1+x^{8}}\right]$

So now, we need to find for $f^{\prime}(1)$

$f^{\prime}(1)=(1+1)\left(1+1^{2}\right)\left(1+1^{4}\right)\left(1+1^{8}\right)\left[\frac{1}{1+1}+\frac{2 \times 1}{1+1^{2}}+\frac{4 \times 1^{3}}{1+1^{4}}+\frac{8 \times 1^{1}}{1+1^{8}}\right]$

$f^{\prime}(1)=(2)(2)(2)(2)\left[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right]$

$f^{\prime}(1)=16\left[\frac{15}{2}\right]$

$=8 \times 15$

$=120$