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

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

A

1

B

0

C

–1

D

None of these

Answer

1

Explanation

Solution

y = (1 + x)(1 + x2)(1 + x4)....(1 + x2nx^{2^{n}})

Multiplying numerator and denominator by (1–x)

⇒ 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)}

⇒ y = (1−x2n+1(1−x))\left( \frac{1 - x^{2^{n + 1}}}{(1 - x)} \right)

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

So  dydx∣x=0\left. \ \frac{dy}{dx} \right|_{x = 0} = −2n+1.0.1+1−012\frac{- 2^{n + 1}.0.1 + 1 - 0}{1^{2}} = 1