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

Question

Dr. Abubacker · Accepted Answer

(c)
$y = \frac{(1 - x)(1 + x)\left( 1 + x^{2} ight)\left( 1 + x^{4} ight)\ldots\ldots..(1 + x^{2n})}{(1 - x)}$
$= \frac{1 - x^{4n}}{1 - x}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx} = \frac{(1 - x)\left( - 4nx^{4n - 1\ } ight)–\left( 1 - x^{4n} ight)( - 1)}{(1 - x)^{2}}$
$= \frac{- 4n(1 - x)x^{4n - 1\ } + \left( 1 - x^{4n} ight)}{(1 - x)^{2}}$
$	herefore\ \ \ \ \ \ \left( \frac{dy}{dx} ight)_{x = 0} = 1$

Q. If $y = (1 + x)\left( 1 + x^{2} \right)\left( 1 + x^{4} \right)\ldots\left( 1 + x^{2n} \right),$ then the value of $\left( \frac{dy}{dx} \right)_{x = 0}$ is

Explanation

Reviewed by Dr. Abubacker

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