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

If y = (1 + x) (1 + x2) (1 + x4)…(1 + x2nx^{2^{n}}) then dydx\frac{dy}{dx} at x = 0 is –

A

1

B

1

C

0

D

None of these

Answer

1

Explanation

Solution

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

∴ dydx\frac{dy}{dx} = –2n+1.x2n+1−1.(1−x)+1−x2n+1(1−x)2\frac{–2^{n + 1}.x^{2^{n + 1} - 1}.(1 - x) + 1 - x^{2^{n + 1}}}{(1 - x)^{2}};

∴ at x = 0, dydx\frac{dy}{dx} = –2n+1.0.1+1−012\frac{–2^{n + 1}.0.1 + 1 - 0}{1^{2}} = 1.