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Question

If y=xnnx1+logxn, then the value of y'(n) is given by


A

1n

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B

1nn

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C

n2+1n

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D

1nnn2+1n

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Solution

The correct option is C

n2+1n


Find the value of y'(n):

The given differential equation is y=xnnx1+logxn

Taking log on both sides,

logy=xnnx1+logxnlogy=nxlogxn+log1+logxn[logab=bloga]

Differentiate the above equation with respect to x.

1yy'=nxnx1n+logxnn+11+logxnnx·1ny'=yn+1x1+logxn+logxnn

Substitute x=n in the above equation and in given equation

y=nnnn1+lognn=1

y'n=1n+1n[log1=0]y'n=n2+1n

Hence, the correct option is C.


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