Question

Find the derivative of the function given by $f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)$ and hence find $f'(1)$

Answer 1

The given relationship is $f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)$
Taking logarithm on both the sides,we obtain
$log f(x)=log(1+x)+log(1+x^2)+log(1+x^4)+log(1+x^8)$
Differentiating both sides with respect to $x$, we obtain
\frac{1}{f(x)}.\frac{d}{dx}[f(x)]=\frac{d}{dx}log(1+x)+\frac{d}{dx}(1+x^2)+\frac{d}{dx}(1+x^4)+\frac{d}{dx}(1+x^8)$$⇒\frac{1}{f(x)}.f'(x)=\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)
$⇒f'(x)=f(x)[\frac{1}{1+x}+\frac{1}{1+x^2}.2x+\frac{1}{1+x^4}.4x^3+\frac{1}{1+x^8}.8x^7]$
$f'(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)[\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}]$
$f'(1)=(1+1)(1+1^2)(1+1^4)(1+1^8)\bigg[\frac{1}{1+1}+\frac{2\times 1}{1+1^2}+\frac{4\times 1^3}{1+1^4}+\frac{8\times 1^7}{1+1^8}\bigg]$
$=2\times2\times2\times2[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}]$
$=16(\frac{1+2+4+8}{2})$
$=16\times \frac{15}{2} =120$