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$y = (4 x + 1) (1 - x^{3}), then \frac{d y}{d x} =$

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$16 x^{3} - 33 x^{2} + 18 x - 1$

${(1 - x)}^{2} (1 - 16 x)$

$- 12 {(1 - x)}^{2}$

$- 16 x^{3} + 33 x^{2} - 18 x + 1$

answer is B, C.

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Detailed Solution

$y = (4 x + 1) {(1 - x)}^{3}$

$\frac{d y}{d x} = (4 x + 1) (3) (1 - x^{2}) (- 1) + {(1 - x)}^{3} (4)$

$= {(1 - x)}^{2} [- 3 (4 x + 1) + 4 (1 - x)]$

$= {(1 - x)}^{2} [- 12 x - 3 + 4 - 4 x]$

$= {(1 - x)}^{2} [1 - 16 x]$

But ${(1 - x)}^{2} (1 - 16 x) = (1 + x^{2} - 2 x) (1 - 16 x)$

$= 1 - 16 x + x^{2} - 16 x^{3} - 2 x + 32 x^{2}$

$= - 16 x^{3} + 33 x^{2} - 18 x + 1$

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