A) \[n!\]
B) \[(n-1)!\]
C) \[{{(-1)}^{n}}(n-1)!\]
D) \[{{(-1)}^{n}}n!\]
Correct Answer: C
Solution :
[c] \[y=\left( 1+\frac{1}{x} \right)\left( 1+\frac{2}{x} \right)\left( 1+\frac{3}{x} \right)...\left( 1+\frac{n}{x} \right)\] \[\frac{dy}{dx}=\left( -\frac{1}{{{x}^{2}}} \right)\left( 1+\frac{2}{x} \right)\left( 1+\frac{3}{x} \right)...\left( 1+\frac{n}{x} \right)\] \[+\left( 1+\frac{1}{x} \right)\left( -\frac{2}{{{x}^{2}}} \right)\left( 1+\frac{3}{x} \right)...\left( 1+\frac{n}{x} \right)\] \[+...+\left( 1+\frac{1}{x} \right)\left( 1+\frac{2}{x} \right)\left( 1+\frac{3}{x} \right)...\left( -\frac{n}{{{x}^{2}}} \right)\] \[\because {{\left. \frac{dy}{dx} \right|}_{x=-1}}=(-1)(-1)(-2)(-3)...(1-n)\] \[={{(-1)}^{n}}(1)(2)(3)...(n-1)={{(-1)}^{n}}(n-1)!\]
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