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

The correct option is B 1 nbsp;y=( 1 x ) ( 1+x ) ( 1+ x ^ 2 ) ...( 1+ x ^ 2 ^ n+1 ) nbsp;/ 1 x = 1 x ^ 2 ^ n+1 / 1 x nbsp;∴ dy / dx = 2 ...

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Properties of Iota

If y=1+x1+x...

Question

If $y = (1 + x) (1 + x^{2}) (1 + x^{4}) . . . . . (1 + x^{2^{n}})$ , find $\frac{d y}{d x}$ at $x = 0$

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Solution

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Similar questions

y = (1 + x) (1 + x^{2}) (1 + x^{4}) . . . (1 + x^{2^{n}})

\frac{d y}{d x}

x = 0

y = (1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}})

\frac{d y}{d x}

y = (1 + x) (1 + x^{2}) (1 + x^{4}) . . . (1 + x^{2^{n}})

\frac{d y}{d x}

x = 0

1

y = \frac{1 - x^{2^{n + 1}}}{1 - x}

y = (1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}})

\frac{d y}{d x}

y = (1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}})

\frac{d y}{d x}

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