If y=x+1x2+1x4+1x8+1 then limx→-1dydx=

Given y=x+1x2+1x4+1x8+1 =x-1x+1x2+1x4+1x8+1x-1 y=x16-1x-1 dydx=x-116x15-x16-1x-12=15x16-16x15+1x-12 limx→-1dydx=limx→-115x16-16x15+1x-12=324=8

If y=x+1x2+1x4+1x8+1 then limx→-1dydx=

Given y=x+1x2+1x4+1x8+1 =x-1x+1x2+1x4+1x8+1x-1 y=x16-1x-1 dydx=x-116x15-x16-1x-12=15x16-16x15+1x-12 limx→-1dydx=limx→-115x16-16x15+1x-12=324=8

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If $y = (x + 1) (x^{2} + 1) (x^{4} + 1) (x^{8} + 1)$ then $\lim_{x \to - 1} \frac{d y}{d x} =$

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answer is D.

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Detailed Solution

$Given y = (x + 1) (x^{2} + 1) (x^{4} + 1) (x^{8} + 1) = \frac{(x - 1) (x + 1) (x^{2} + 1) (x^{4} + 1) (x^{8} + 1)}{x - 1}$
$y = \frac{(x^{16} - 1)}{x - 1} \frac{d y}{d x} = \frac{(x - 1) 16 x^{15} - (x^{16} - 1)}{{(x - 1)}^{2}} = \frac{15 x^{16} - 16 x^{15} + 1}{{(x - 1)}^{2}}$ $\lim_{x \to - 1} \frac{d y}{d x} = \lim_{x \to - 1} \frac{15 x^{16} - 16 x^{15} + 1}{{(x - 1)}^{2}} = \frac{32}{4} = 8$