Find the derivative of the function given by

The given relationship is $f(x)=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)$

Taking logarithm on both the sides, we obtain

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

Differentiating both sides with respect to x, we obtain

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

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

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

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

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

$=2 \times 2 \times 2 \times 2\left[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right]$

$=16 \times\left(\frac{1+2+4+8}{2}\right)$

$=16 \times \frac{15}{2}=120$