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
\(\frac{1}{f(x)}.\frac{d}{dx}[f(x)]=\frac{d}{dx}log(1+x)+\frac{d}{dx}(1+x^2)+\frac{d}{dx}(1+x^4)+\frac{d}{dx}(1+x^8)\)\(⇒\frac{1}{f(x)}.f'(x)=\frac{1}{1+x}.\frac{d}{dx}(1+x)+\frac{1}{1+x^2}\frac{d}{dx}(1+x^2)+\frac{1}{1+x^4}\frac{d}{dx}(1+x^4)+\frac{1}{1+x^8}\frac{d}{dx}(1+x^8)\)
\(⇒f'(x)=f(x)[\frac{1}{1+x}+\frac{1}{1+x^2}.2x+\frac{1}{1+x^4}.4x^3+\frac{1}{1+x^8}.8x^7]\)
\(f'(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)[\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}]\)
\(f'(1)=(1+1)(1+1^2)(1+1^4)(1+1^8)\bigg[\frac{1}{1+1}+\frac{2\times 1}{1+1^2}+\frac{4\times 1^3}{1+1^4}+\frac{8\times 1^7}{1+1^8}\bigg]\)
\(=2\times2\times2\times2[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}]\)
\(=16(\frac{1+2+4+8}{2})\)
\(=16\times \frac{15}{2} =120\)