IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

juana

$y={e}^{{x+e}^{x.....}}$ till infinity

ashish_banga

y/1-y

juana

Pls tell me how you got it. Should we assume y=t.
So, y=e^t and the take log and differentiate?

nasa_hs

y=e^x+y

log y = x+y

after differentiating

ans is dy/dx=y/1-y

ashish_banga

nasa, check again ur answer

nasa_hs

The assumption y=e^x+ywill not affect the final ans cause u need to find dy/dx till infinty

Thanks ashish banga mistake corrected

abhay_yadav80

nasa is wrong

abhay_yadav80

no sorry yaar u r absolutly right

ans. is y/1-y

hats off to u

juana

Thanks nasa. I was assuming it as t instead of y itself