Question

Find the derivative of the function given by $f\left(x\right)=(1+x)(1+x^2)(1+x^4)(1+x^8)$ and hence find $f^{\prime }\left(1\right).$

Answer 1

The given equation is $f\left(x\right)=(1+x)(1+x^2)(1+x^4)(1+x^8)$
Taking logarithm on both the sides, we obtain
$\log f\left(x\right)=\log (1+x)+\log (1+x^2)+\log (1+x^4)+\log (1+x^8)$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{f\left(x\right)}.\frac{d}{dx}\left[f\right(x\left)\right]=\frac{d}{dx}\log (1+x)+\frac{d}{dx}\log (1+x^2)+\frac{d}{dx}\log (1+x^4)+\frac{d}{dx}\log (1+x^8)$
$\Rightarrow \frac{1}{f\left(x\right)}.f^{\prime }\left(x\right)=\frac{1}{1+x}.\frac{d}{dx}(1+x)+\frac{1}{1+x^2}.\frac{d}{dx}(1+x^2)+\frac{1}{1+x^4}.\frac{d}{dx}(1+x^4)+\frac{1}{1+x^8}.\frac{d}{dx}(1+x^8)$
$\Rightarrow f^{\prime }\left(x\right)=f\left(x\right)[\frac{1}{1+x}+\frac{1}{1+x^2}.2x+\frac{1}{1+x^4}.4x^3+\frac{1}{1+x^8}.8x^7]$
$\therefore f^{\prime }\left(x\right)=(1+x)(1+x^2)(1+x^4)(1+x^8)[\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}]$
Hence, $f^{\prime }\left(1\right)=(1+1)(1+1^2)(1+1^4)(1+1^8)[\frac{1}{1+1}+\frac{2\times 1}{1+1^2}+\frac{4\times 1^3}{1+1^4}+\frac{8\times 1^7}{1+1^8}]$
$=2\times 2\times 2\times 2[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}]$
$=16\times \left(\frac{1+2+4+8}{2}\right)$
$=16\times \frac{15}{2}=120$