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Question

# If y=x(logx)log(logx), then dydx is

A
yx[lnxlnx1+2lnxln(lnx)]
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B
yxlnx[(lnx)2+2ln(lnx)]
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C
yx(logx)log(logx)(2log(logx)+1)
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D
ylogyxlogx[2log(logx)+1]
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Solution

## The correct option is D ylogyxlogx[2log(logx)+1]y=x(logx)log(logx) Taking log on both side, logy=(logx)(logx)log(logx)⋯(1) Taking log on both side, log(logy)=log(logx)+log(logx)log(logx) Differentiating w.r.t. x, 1logy⋅1y⋅dydx=1xlogx+2log(logx)logx⋅1x⇒1logy⋅1y⋅dydx=2log(logx)+1xlogx⇒dydx=yx⋅logylogx(2log(logx)+1) Using equation (1), dydx=yx(logx)log(logx)(2log(logx)+1)

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