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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,
1logy1ydydx=1xlogx+2log(logx)logx1x1logy1ydydx=2log(logx)+1xlogxdydx=yxlogylogx(2log(logx)+1)
Using equation (1),
dydx=yx(logx)log(logx)(2log(logx)+1)

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