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

Answer: (A)

$1$

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

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

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

$.........$
$=\frac{1-x^{2^{n+1}}}{1-x}$
$\Rightarrow \frac{dy}{dx}=\frac{-2^{n+1}{x}^{2^{n+1}-1}(1-x)+(1-x^{2^{n+1}})}{{(1-x)}^2}$
$\frac{dy}{dx}{|}_{\text{at}x=0}=1$