Question

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

• A. $0$
• B. $(-1)(n-1)!$
• C. $n!-1$
• D. ${\left(-1\right)}^{n-1}\left(n-1\right)!$

Answer 1

Answer: (B)

$\because y=\left(1-x\right)\left(2-x\right)\dots \dots \left(n-x\right)$
Taking log on both sides, we get
$logy=log\left(1-x\right)+log\left(2-x\right)+...+log\left(n-x\right)$
$\frac{1}{y}\frac{dy}{dx}=\frac{1}{\left(1-x\right)}\left(-1\right)+\frac{1}{\left(2-x\right)}(-1)+\dots \dots +\frac{1}{\left(n-x\right)}\left(-1\right)$
$\frac{dy}{dx}=-y\left[\frac{\left[\left(2-x\right)\left(3-x\right)\dots \dots \left(n-x\right)+\dots \dots \right]}{y}\right]$
$\therefore {\left(\frac{dy}{dx}\right)}_{x=1}=1\cdot 2\dots \dots \left(n-1\right)\left(-1\right)$
$=\left(-1\right)\left(n-1\right)!$