Differentiation Question 122
Question: If $ y=( 1+\frac{1}{x} )( 1+\frac{2}{x} )( 1+\frac{3}{x} )….( 1+\frac{n}{x} ) $ and $ x\ne 0. $ then $ \frac{dy}{dx} $ when $ x=-1 $ is
Options:
A) $ n! $
B) $ (n-1)! $
C) $ {{(-1)}^{n}}(n-1)! $
D) $ {{(-1)}^{n}}n! $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ y=( 1+\frac{1}{x} )( 1+\frac{2}{x} )( 1+\frac{3}{x} )…( 1+\frac{n}{x} ) $
$ \frac{dy}{dx}=( -\frac{1}{x^{2}} )( 1+\frac{2}{x} )( 1+\frac{3}{x} )…( 1+\frac{n}{x} ) $
$ +( 1+\frac{1}{x} )( -\frac{2}{x^{2}} )( 1+\frac{3}{x} )…( 1+\frac{n}{x} ) $
$ +…+( 1+\frac{1}{x} )( 1+\frac{2}{x} )( 1+\frac{3}{x} )…( -\frac{n}{x^{2}} ) $
$ \because {{. \frac{dy}{dx} |} _{x=-1}}=(-1)(-1)(-2)(-3)…(1-n) $
$ ={{(-1)}^{n}}(1)(2)(3)…(n-1)={{(-1)}^{n}}(n-1)! $
BETA
