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

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(C) Given $y=(1+x)(1+x^2)(1+x^4) \dots (1+x^{2n})$.Taking $\log$ on both sides,we get $\log y = \log(1+x) + \log(1+x^2) + \log(1+x^4) + \dots + \log(1

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$1$

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(C) Given $y=(1+x)(1+x^2)(1+x^4) \dots (1+x^{2n})$.Taking $\log$ on both sides,we get $\log y = \log(1+x) + \log(1+x^2) + \log(1+x^4) + \dots + \log(1+x^{2n})$.Differentiating both sides with respect to $x$,we get $\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \dots + \frac{2n \cdot x^{2n-1}}{1+x^{2n}}$.At $x=0$,the value of $y = (1+0)(1+0) \dots (1+0) = 1$.Substituting $x=0$ and $y=1$ in the derivative equation,we get $\frac{1}{1} \left. \frac{dy}{dx} \right|_{x=0} = \frac{1}{1+0} + 0 + 0 + \dots + 0$.Therefore,$\left. \frac{dy}{dx} \right|_{x=0} = 1$.

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

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