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
$ log\text{ }y=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}}=1.2....(n-1)(-1) $ $ =(-1)(n-1)! $