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

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Question

(B) Let $y = f(x) = (1 - x)(2 - x)(3 - x)...(n - x)$.Using the product rule for differentiation,the derivative is given by:$\frac{dy}{dx} = \frac{d}{d

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $y = f(x) = (1 - x)(2 - x)(3 - x)...(n - x)$.Using the product rule for differentiation,the derivative is given by:$\frac{dy}{dx} = \frac{d}{dx}[(1 - x)] \cdot (2 - x)(3 - x)...(n - x) + (1 - x) \cdot \frac{d}{dx}[(2 - x)(3 - x)...(n - x)]$.Since $\frac{d}{dx}(1 - x) = -1$,we have:$\frac{dy}{dx} = -1 \cdot (2 - x)(3 - x)...(n - x) + (1 - x) \cdot \frac{d}{dx}[(2 - x)(3 - x)...(n - x)]$.Now,evaluate at $x = 1$:$\left. \frac{dy}{dx} \right|_{x=1} = -1 \cdot (2 - 1)(3 - 1)...(n - 1) + (1 - 1) \cdot [\dots]$.$\left. \frac{dy}{dx} \right|_{x=1} = -1 \cdot (1)(2)(3)...(n - 1) + 0$.$\left. \frac{dy}{dx} \right|_{x=1} = -1 \cdot (n - 1)!$.This can be written as $(-1)^1 (n - 1)!$. However,looking at the product $(1-x)(2-x)...(n-x)$,there are $n$ terms. The derivative at $x=1$ is $-(2-1)(3-1)...(n-1) = -(1)(2)...(n-1) = -(n-1)!$. Thus,the correct option is $B$.

The derivative of $y = (1 - x)(2 - x)...(n - x)$ at $x = 1$ is equal to

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