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January 30, 2015

January 30, 2015

Posted by **Abhishek** on Monday, June 4, 2012 at 7:25am.

- Calculus -
**Steve**, Monday, June 4, 2012 at 10:29amy = √(lnx+√(lnx+√(lnx+...)))

y^2 = lnx+√(lnx+√(lnx+...))) = lnx + y

2y y' = 1/x + y'

(2y-1)y' = 1/x

y' = 1/(x(2y-1))

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