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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 Jun 2008 20:35:51 IST
Accepted Answer [?]
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taking los to the base e both the sides
logx^y =log[ e^(x-y)]
y logx = x-y
differentiatintg both sides
dy/dxlogx+ Y/x = 1 - dy/dx
dy.dx ( logx+1) = 1-y/x
= [ 1-y/x] / logx+1
but ylogx =x-y
y ( logx+1) = x
y = x / (logx+1)
so dy/dx = [ 1- x / (logx+1)x ] / (logx+1)
= logx +1 -1 / (logx+1)^2
= logx / (logx+1)^2
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this reply: 15 points
(with 3 
in 3 votes ) [?]
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