If x = e^(y + e(y + e^(y + ..... oo) then dy/dx is ??

2 Answers
Aug 14, 2017

y' = 1/x-1

Explanation:

Applying log to both sides

log x = y + e^(y cdots) Aplying again

log(logx-y) = y + e^(y cdots) so

logx = log(log x-y) or

x = logx - y now deriving both sides

1 = 1/x - y' and then

y' = 1/x-1

Aug 14, 2017

dy/dx=(1-ex)/(x(1+e))

Explanation:

Assuming I correctly interpreted the problem:

x=e^(y+e(y+e^(y+e(y+e^(y+...))))

lnx=y+e(y+e^(y+e(y+e^(y+...))))

lnx=y+e(y+x)

y=(lnx-ex)/(1+e)

dy/dx=(1//x-e)/(1+e)=(1-ex)/(x(1+e))