Question

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

Answer 1

`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))`

Answer 2

`(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))`