The derivative of y = (1-x)(2-x)(3-x) dots (n | vedclass

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(C) Let $y = f(x) = (1-x)(2-x)(3-x) \dots (n-x)$.Using the product rule for differentiation,if $y = u_1 u_2 u_3 \dots u_n$,then $\frac{dy}{dx} = u_1'

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

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(C) Let $y = f(x) = (1-x)(2-x)(3-x) \dots (n-x)$.Using the product rule for differentiation,if $y = u_1 u_2 u_3 \dots u_n$,then $\frac{dy}{dx} = u_1' (u_2 u_3 \dots u_n) + u_1 (u_2 u_3 \dots u_n)'$.At $x=1$,the term $(1-x)$ becomes $0$.Therefore,in the derivative,all terms containing $(1-x)$ will vanish except for the term where $(1-x)$ is differentiated.Let $g(x) = (2-x)(3-x) \dots (n-x)$. Then $y = (1-x)g(x)$.$\frac{dy}{dx} = (-1)g(x) + (1-x)g'(x)$.At $x=1$,$\frac{dy}{dx} = (-1)g(1) + 0 = -g(1)$.$g(1) = (2-1)(3-1)(4-1) \dots (n-1) = (1)(2)(3) \dots (n-1) = (n-1)!$.Thus,$\frac{dy}{dx} 	ext{ at } x=1 = -(n-1)!$.

The derivative of $y = (1-x)(2-x)(3-x) \dots (n-x)$ at $x=1$ is

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