The derivative of y=(1−x)(2−x)…(n−x) at x=1 is

dydx=−[(2−x)(3−x)…(n−x)+(1−x)(3−x)…(n−x)+…(1−x)(2−x)…(n−1−x)]⇒dydxx=1=(−1)(1)(2)…(n−1)=(−1)(n−1) !

The derivative of y=(1−x)(2−x)…(n−x) at x=1 is

dydx=−[(2−x)(3−x)…(n−x)+(1−x)(3−x)…(n−x)+…(1−x)(2−x)…(n−1−x)]⇒dydxx=1=(−1)(1)(2)…(n−1)=(−1)(n−1) !

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The derivative of $y = (1 - x) (2 - x) \dots (n - x) at x = 1$ is

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$(- 1) (n - 1)!$

$n! - 1$

$(- 1)^{n - 1} (n - 1)!$

answer is B.

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Detailed Solution

$\frac{d y}{d x} = - [(2 - x) (3 - x) \dots (n - x) + (1 - x) (3 - x)$ $\dots (n - x) + \dots (1 - x) (2 - x) \dots (n - 1 - x)]$ $\Rightarrow {(\frac{d y}{d x})}_{x = 1} = (- 1) (1) (2) \dots (n - 1) = (- 1) (n - 1)!$

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The derivative of $y = (1 - x) (2 - x) \dots (n - x) at x = 1$ is