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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
(A) 0
(B) \(-1\)
(C) 1
(D) 2

3 Answers

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Ans: (C)
Hint : \(\quad y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots .\left(1+x^{2 n}\right)\)
$$
\begin{aligned}
&\Rightarrow \ln y=\ln (1+x)+\ln \left(1+x^{2}\right) \ldots+\ln \left(1+x^{2 n}\right) \\
&\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{3 x^{2}}{1+x^{3}}+\ldots .+\frac{2 n x^{2 n-1}}{\left(1+x^{2 n}\right)} \\
&\Rightarrow \frac{d y}{d x}=y \cdot\left(\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{3 x^{2}}{1+x^{3}}+\ldots . .+\frac{2 n x^{2 n-1}}{1+x^{2 n}}\right) \\
&\Rightarrow\left|\frac{d y}{d x}\right|_{x=0}=1.1=1
\end{aligned}
$$
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