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

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Answer 1 · 2017-08-14T16:39:50

#y' = 1/x-1#

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#

Answer 2 · 2017-08-14T16:42:01

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

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))#