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

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

Aicharya · Accepted Answer

$y=(1+x)\left(1+x^2ight)\left(1+x^4ight) \ldots\left(1+x^{2 n}ight)$ Taking ' $\log$ ' on both sides, we get $\begin{aligned} \log y=\log (1+x)+\log \left(1+x^2ight) & +\log \left(1+x^4ight) \ & +\ldots+\log \left(1+x^{2 \mathrm{n}}ight) \end{aligned}$ Differentiating w.r.t. $x$, we get $\frac{1}{y} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{1+x}+\frac{2 x}{1+x^2}+\frac{4 x^3}{1+x^4}+\ldots+\frac{2 \mathrm{n} 	imes x^{2 \mathrm{n}-1}}{1+x^{2 \mathrm{n}}}$ At $x=0$, (i) $\Rightarrow y=1$ $	herefore \quad\left(ight.$ ii) $\left.\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}ight|_{x=0}=1+0+0+\ldots+0=1$

$y=(1+x)\left(1+x^2\right)\left(1+x^4\right) \ldots \ldots \ldots\left(1+x^{2 n}\right) \text {, }$ then the…

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