Differentiate y=a^x^a^x^a----to infinity?

2 Answers
Sep 2, 2017

#dy/dx=(y^2lny)/(x(1-ylnxlny))#

where #y=a^(x^(a^(x^cdots)))# and #lny=x^(a^(x^(a^cdots)))lna#

Explanation:

#y=a^(x^(a^(x^cdots)))#

Take the natural logarithm:

#lny=ln(a^(x^(a^(x^cdots))))#

#color(white)lny=x^(a^(x^(a^cdots)))lna#

Take the natural logarithm once more:

#ln(lny)=ln(x^(a^(x^(a^cdots)))lna)#

#color(white)ln(lny)=ln(x^(a^(x^(a^cdots))))+ln(lna)#

#color(white)ln(lny)=a^(x^(a^(x^cdots)))lnx+ln(lna)#

#color(white)ln(lny)=ylnx+ln(lna)#

Taking the derivative of this last version:

#d/dxln(lny)=d/dx(ylnx)+d/dxln(lna)#

Using the chain rule (left) and product rule (right):

#1/lny(d/dxlny)=dy/dxlnx+y/x+0#

#1/lny(1/y)dy/dx=dy/dxlnx+y/x#

Grouping #dy/dx# terms:

#dy/dx(1/(ylny)-lnx)=y/x#

#dy/dx((1-ylnxlny)/(ylny))=y/x#

#dy/dx=(y^2lny)/(x(1-ylnxlny))#

Sep 9, 2017

#(dy)/(dx) = (y^2 Log y)/(x(1 - y Logx Log y))#

Explanation:

If #y = a^(x^(a^(x^cdots)))#

then

#y = a^(x^y)# then applying #log# to both sides

#log y = x^y log(a)# and now applying again #log# to both sides

#log(logy) = y log x+log(loga)# and now deriving

#(dy)/(dx) =(y^2 Log y)/(x(1 - y Logx Log y))#