If y-1-x1-x21-x4- 1-x2n then dy-dx at x - 0 is

Since, y=(1+x)(1+x^2)(1+x^4)⋯ (1+x^2^n), (1 x)y=(1 x^2)(1+x^2)(1+x^4)⋯ (1+x^2^n) =(1 x^4)(1+x^4)⋯ (1+x^2^n+1) ∴ y=1 x^2^n+1/(1 x) ∵dy/dx = (1 x)( 2^n+1.x^2^n ...

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Byju's Answer

Standard XI

Mathematics

Quotient Rule of Differentiation

If y=1+x1+x21...

Question

If $y = (1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}})$ then $\frac{d y}{d x}$ at x = 0 is

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Solution

The correct option is C 1
$S i n c e, y = (1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}}), (1 - x) y = (1 - x^{2}) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}}) = (1 - x^{4}) (1 + x^{4}) \dots (1 + x^{2^{n + 1}}) ∴ y = \frac{1 - x^{2^{n + 1}}}{(1 - x)} ∵ \frac{d y}{d x} = \frac{(1 - x) (- 2^{n + 1} . x^{2^{n}}) - (1 - x^{2^{n + 1}}) (- 1)}{(1 - x)^{2}} ∴ {\frac{d y}{d x} ∣ ∣}_{x = 0} = \frac{(1) (0) - (1) (- 1)}{(1)^{2}} = 1$

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If y=(1+x)(1+x2)(1+x4)⋯(1+x2n) then dydx at x = 0 is

The correct option is C 1Since, y=(1+x)(1+x2)(1+x4)⋯(1+x2n),(1−x)y=(1−x2)(1+x2)(1+x4)⋯(1+x2n)=(1−x4)(1+x4)⋯(1+x2n+1)∴y=1−x2n+1(1−x)∵dydx=(1−x)(−2n+1.x2n)−(1−x2n+1)(−1)(1−x)2∴dydx∣∣x=0=(1)(0)−(1)(−1)(1)2=1

If $y = (1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}})$ then $\frac{d y}{d x}$ at x = 0 is