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Question

If xloge(logex)x2+y2=4 (y>0), then dydx at x=e is equal to :

A
(2e1)24+e2
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B
e4+e2
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C
(1+2e)24+e2
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D
(1+2e)4+e2
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Solution

The correct option is A (2e1)24+e2
y2=4+x2xloge(logex) (1)
For x=e
y2=4+e2e×0
y=4+e2

Differentiate equation (1) w.r.t. x, we get
2ydydx=2xloge(logex)x×1logex×1x

dydx=2xloge(logex)1logex2y

dydxx=e=2e124+e2

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