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If $y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots\left(1+x^{2 n}\right),$ then the value of $\left(\frac{d y}{d x}\right)$ at $x=0$ is
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Given,
$y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)+\ldots+\left(1+x^{2 n}\right)$
$\Rightarrow \log y=\log (1+x)+\log \left(1+x^{2}\right)$ $+\log \left(1+x^{4}\right)+\ldots+\log \left(1+x^{2n}\right)$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{3 x^{2}}{1+x^{3}}+\ldots+\frac{2 n x^{2 n-1}}{1+x^{2 n}}$
$\Rightarrow \frac{d y}{d x}=y\left[\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{3 x^{2}}{1+x^{3}}+\ldots+\frac{2 n x^{2 n-1}}{1+x^{2 n}}\right]$
$\therefore \quad\left(\frac{d y}{d x}\right)_{x=0}=y\left(\frac{1}{1+0}\right)=1$
$y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)+\ldots+\left(1+x^{2 n}\right)$
$\Rightarrow \log y=\log (1+x)+\log \left(1+x^{2}\right)$ $+\log \left(1+x^{4}\right)+\ldots+\log \left(1+x^{2n}\right)$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{3 x^{2}}{1+x^{3}}+\ldots+\frac{2 n x^{2 n-1}}{1+x^{2 n}}$
$\Rightarrow \frac{d y}{d x}=y\left[\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{3 x^{2}}{1+x^{3}}+\ldots+\frac{2 n x^{2 n-1}}{1+x^{2 n}}\right]$
$\therefore \quad\left(\frac{d y}{d x}\right)_{x=0}=y\left(\frac{1}{1+0}\right)=1$
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