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Question

If y = xlog xlog log x,then dydx=?


A

xlogxloglogx1xlogxlogx+loglogx+loglogx1x+1xlogx

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B

xlogxloglogxlog (log x)2logx+1x

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C

xlogxxlogxloglogxx1logx+1

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D

ylogyxlogx2loglogx+1

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Solution

The correct option is D

ylogyxlogx2loglogx+1


Explanation for the correct option:

Step 1: Simplifying the given equation:

Given that,

y = xlogxlog log x

By taking log on both sides,

logy=logxlogxloglogx =logxloglogx.logx [ logab=bloga]

Now, by taking log again on both sides,

loglogy=loglogxloglogx.logx =loglogxloglogx+loglogx [ log(ab)=loga+logb] =loglogx.loglogx+loglogx=loglogx2+loglogx ...1

Step 2: Find the dydx:

Differentiate equation 1 both sides with respect to x, we get

1logy.1ydydx=2loglogx.1logx.1x+1logx×1x [ ddx(logx)=1x] 1ylogy.dydx=1xlogx2loglogx+1 dydx=ylogyxlogx2loglogx+1

Hence, the correct option is D.


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