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Question: If y = (1 + x) (1 + x<sup>2</sup>) (1 + x<sup>4</sup>) …. (1 + x<sup>2n</sup>) then \(\frac{dy}{dx}\...

If y = (1 + x) (1 + x2) (1 + x4) …. (1 + x2n) then dydx\frac{dy}{dx} at x = 0 is

A

1

B

– 1

C

0

D

None

Answer

1

Explanation

Solution

y = (1 + x) (1 + x2) (1 + x4) …….(1 + x2n)

Ž y =(1−x)(1+x)(1+x2)......(1+x2n)(1−x)=1−x2n+11−x\frac{(1 - x)(1 + x)(1 + x^{2})......(1 + x^{2n})}{(1 - x)} = \frac{1 - x^{2n + 1}}{1 - x}

dydx=(1−x){0−(2n+1).x2n−(1−x)2n(0−1)}(1−x)2⇒dydx(x=0)=1\frac{dy}{dx} = \frac{(1 - x)\left\{ 0 - (2n + 1).x^{2n} - (1 - x)^{2n}(0 - 1) \right\}}{(1 - x)^{2}} \Rightarrow \frac{dy}{dx}_{(x = 0)} = 1