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

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(C) Given,$y=(1+x)(1+x^{2})(1+x^{4}) \ldots (1+x^{2^{n}})$.Taking the natural logarithm on both sides:$\log y = \log(1+x) + \log(1+x^{2}) + \log(1+x^{

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(C) Given,$y=(1+x)(1+x^{2})(1+x^{4}) \ldots (1+x^{2^{n}})$.Taking the natural logarithm on both sides:$\log y = \log(1+x) + \log(1+x^{2}) + \log(1+x^{4}) + \ldots + \log(1+x^{2^{n}})$.Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{d y}{d x} = \frac{1}{1+x} + \frac{2x}{1+x^{2}} + \frac{4x^{3}}{1+x^{4}} + \ldots + \frac{2^{n}x^{2^{n}-1}}{1+x^{2^{n}}}$.Thus,$\frac{d y}{d x} = y \left[ \frac{1}{1+x} + \frac{2x}{1+x^{2}} + \ldots + \frac{2^{n}x^{2^{n}-1}}{1+x^{2^{n}}} ight]$.At $x=0$,$y = (1+0)(1+0) \ldots (1+0) = 1$.Substituting $x=0$ in the derivative expression:$\left(\frac{d y}{d x}ight)_{x=0} = 1 \left[ \frac{1}{1+0} + 0 + 0 + \ldots + 0 ight] = 1 	imes 1 = 1$.

List-$I$	List-$II$
$A$. $\sec^{-1} x$	$I$. $\frac{1}{1-x^2}, x \in (-1, 1)$
$B$. $\tanh^{-1} x$	$II$. $\frac{-1}{\|x\| \sqrt{x^2+1}}, x \neq 0$
$C$. $\coth^{-1} x$	$III$. $\frac{1}{\|x\| \sqrt{x^2-1}}, \|x\| > 1$
$D$. $\operatorname{cosech}^{-1} x$	$IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$
	$V$. $\frac{-1}{\|x\| \sqrt{1-x^2}}, \|x\| < 1, x \neq 0$

If $y=(1+x)(1+x^{2})(1+x^{4}) \ldots (1+x^{2^{n}}),$ then the value of $\left(\frac{d y}{d x}\right)$ at $x=0$ is

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