Question

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

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

Answer 1

Answer: (B)

$\because$ $y=(1-x)(2-x)....(n-x)$
On taking log on both sides, we get
$logy=log(1-x)+log(2-x)+...$ $+\log (n-x)$
$\frac{1}{y}\frac{dy}{dx}=\frac{1}{(1-x)}(-1)+\frac{1}{(2-x)}(-1)+....$
$+\frac{1}{(n-x)}(-1)$ $\frac{dy}{dx}=y\left[\frac{(2-x)(3-x).....(n-x)(-1)+.....}{y}\right]$
${\left(\frac{dy}{dx}\right)}_{x=1}=\mathrm{1.2....}(n-1)(-1)$ $=(-1)(n-1)!$