Question

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

• A. -1
• B. 1
• C. None of these

Answer 1

The correct option is B 1

$Since,y=(1+x)(1+x^2)(1+x^4)\cdots (1+x^{2^n}),(1-x)y=(1-x^2)(1+x^2)(1+x^4)\cdots (1+x^{2^n})=(1-x^4)(1+x^4)\cdots (1+x^{2^{n+1}})\therefore y=\frac{1-x^{2^{n+1}}}{(1-x)}\because \frac{dy}{dx}=\frac{(1-x)(-2^{n+1}.x^{2^n})-(1-x^{2^{n+1}})(-1)}{(1-x{)}^2}\therefore {\frac{dy}{dx}|}_{x=0}=\frac{\left(1\right)\left(0\right)-\left(1\right)(-1)}{(1{)}^2}=1$