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Question

If y=loge(x+loge(x+....)), then dydx at (x=e22,y=2) is

A
1e21
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B
log222(e21)
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C
2loge2(e21)
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D
None of these
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Solution

The correct option is B 1e21
Given,

loge(x+log(x+)e)

y=logx+ye (as goes on to infinity)

ey=x+y

Differentiate on both sides

eydydx=1+dydx

dydx=1ey1

at(x=e22,y=2)=1e21


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