If \[y=\left( 1+\frac{1}{x} \right)\left( 1+\frac{2}{x} \right)\left( 1+\frac{3}{x} \right)....\left( 1+\frac{n}{x} \right)\] and \[x\ne 0.\] then \[\frac{dy}{dx}\] when \[x=-1\] is

If \[y=\left( 1+\frac{1}{x} \right)\left( 1+\frac{2}{x} \right)\left(

1.	If \[y=\left( 1+\frac{1}{x} \right)\left( 1+\frac{2}{x} \right)\left( 1+\frac{3}{x} \right)....\left( 1+\frac{n}{x} \right)\] and \[x\ne 0.\] then \[\frac{dy}{dx}\] when \[x=-1\] is
A.	\[n!\]
B.	\[(n-1)!\]
C.	\[{{(-1)}^{n}}(n-1)!\]
D.	\[{{(-1)}^{n}}n!\]
Answer» D. \[{{(-1)}^{n}}n!\]

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If \[y=\left( 1+\frac{1}{x} \right)\left( 1+\frac{2}{x} \right)\left( 1+\frac{3}{x} \right)....\left( 1+\frac{n}{x} \right)\] and \[x\ne 0.\] then \[\frac{dy}{dx}\] when \[x=-1\] is

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