If y=x(logx)log(logx), then dydxis

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yx(1n xlogx−1)+2 1n x 1n (1nx)

yx(logx)log(logx)(2log(logx)+1)

yx 1nx[(1nx)2+21n(1nx)]

yxlogylogx[2log(logx)+1]

answer is B.

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Detailed Solution

y=x(logx)log(logx)⇒log y=(logx)(logx)log(logx)→(1)Taking log of both sides, we get ⇒log (logy)=log(logx)+log(logx)log(logx)Diff. w.r.t x, we get 1logy.1ydydx=1xlogx+2log(logx)logx1x=2log(logx)+1xlogx⇒dydx=yx.logylogx(2log(logx)+1)Substituting the value of log y from (1), we get yx(logx)log(logx)(2log(logx)+1)

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