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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

If $f(t) = \frac{t}{2} + \frac{1}{4} \log(2t - 1)$,then $f^{\prime}\left(\frac{t+1}{2t+1}\right) = $

If $f(x) = 1 + x + x^2 + \ldots + x^{1000}$,then $f^{\prime}(-1) = $ . . . . . .

If $f(x) = \frac{x}{1+|x|}$,for $x \in \mathbb{R}$,then $f^{\prime}(0)$ is equal to :

If $y=a \sin x+b \cos x$,then $y^2+\left(\frac{d y}{d x}\right)^2$ is a

$\frac{d}{dx} [\operatorname{cosech}^{-1}(\tan 2x)] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module