Question

If $y=(1+x)(1+x^2)(1+x^4)....(1+x^{2^n})$, then ${\left(\frac{dy}{dx}\right)}_{x=0}=$

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

Answer: (C)

$1$

$y(1-x)=(1-x)(1+x)(1+x^2)(1+x^4)....(1+x^{2^n})$, ........Multiply both sides by $(1-x)$

$y(1-x)=(1-x^2)(1+x^2)(1+x^4)....(1+x^{2^n})$

$\Rightarrow y(1-x)=(1-x^{2^n})(1+x^{2^n})=1-x^{2^{n+1}}$, Use $(a+b)(a-b)=a^2-b^2$

$\Rightarrow (1-x)\frac{dy}{dx}-y=-2^{n+1}{x}^{2^{n+1}-1}$, Differentiate both sides

$\therefore {\left(\frac{dy}{dx}\right)}_{x=0}=y{|}_{x=0}=1$