Question

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

• A. $1$
• B. $-1$
• C. $0$
• D. none of these

Answer 1

The correct option is A $1$

$y=(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})$
Multiply and dividing by (1-x)
$y=\frac{(1-x)(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})}{(1-x)}$
$y=\frac{(1-x^{2^{n+1}})}{(1-x)}$
$\frac{dy}{dx}=\frac{(1-x)(-2^{n+1}{x}^{2^{n+1}-1})-(1-x^{2^{n+1}})(-1)}{{(1-x)}^2}$
${\left(\frac{dy}{dx}\right)}_{x=0}=1$