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Question

The derivative of y=(1x)(2x)...(nx) at x=1 is

A
0
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B
(n1)!
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C
n!1
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D
(1)n1(n1)!
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Solution

The correct option is B (n1)!
y=(1x)(2x)...(nx)dydx=[(2x)...(nx)+(1x)(3x)...(nx) +...(1x)(2x)...(n1x)]
Putting x=1
dydx=(1)(2)...(n1)dydx=(n1)!

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