Question

If $f\left(x\right)=(1+x)(1+x^2)(1+x^4)(1+x^8),$ then the value of $f^{\prime }\left(1\right)$ is

Answer 1

Given, $f\left(x\right)=(1+x)(1+x^2)(1+x^4)(1+x^8)$
Taking ${\log }_e$ on both sides,
$\Rightarrow \ln f\left(x\right)=\ln (1+x)+\ln (1+x^2)+\ln (1+x^4)+\ln (1+x^8)$
Differentiating w.r.t. $x,$ we get
$\frac{1}{f\left(x\right)}\cdot f^{\prime }\left(x\right)=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}$
$\Rightarrow f^{\prime }\left(x\right)=f\left(x\right)[\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}]$
Hence, $f^{\prime }\left(1\right)=f\left(1\right)[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}]$
$=(2{)}^4\times \frac{15}{2}=120$