If y=(1+x)(1+x^2)(1+x^4) dots (1+x^{2^n}),the | vedclass

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(C) Given $y = (1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})$.Multiply and divide by $(1-x)$:$y = \frac{(1-x)(1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})}{1-x}$Using

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(C) Given $y = (1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})$.Multiply and divide by $(1-x)$:$y = \frac{(1-x)(1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})}{1-x}$Using the identity $(a-b)(a+b) = a^2-b^2$ repeatedly:$y = \frac{(1-x^2)(1+x^2)(1+x^4) \dots (1+x^{2^n})}{1-x} = \frac{(1-x^4)(1+x^4) \dots (1+x^{2^n})}{1-x} = \dots = \frac{1-x^{2^{n+1}}}{1-x}$Now,differentiate using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{vu' - uv'}{v^2}$:$\frac{dy}{dx} = \frac{(1-x)(-2^{n+1}x^{2^{n+1}-1}) - (1-x^{2^{n+1}})(-1)}{(1-x)^2}$At $x=0$:$(\frac{dy}{dx})_{x=0} = \frac{(1-0)(0) - (1-0)(-1)}{(1-0)^2} = \frac{0 + 1}{1} = 1$.

If $y=(1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})$,then $(\frac{dy}{dx})_{x=0}$ is equal to

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