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

Sep 2, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{2} \ln y}{x \left(1 - y \ln x \ln y\right)}$

where $y = {a}^{{x}^{{a}^{{x}^{\cdots}}}}$ and $\ln y = {x}^{{a}^{{x}^{{a}^{\cdots}}}} \ln a$

#### Explanation:

$y = {a}^{{x}^{{a}^{{x}^{\cdots}}}}$

Take the natural logarithm:

$\ln y = \ln \left({a}^{{x}^{{a}^{{x}^{\cdots}}}}\right)$

$\textcolor{w h i t e}{\ln} y = {x}^{{a}^{{x}^{{a}^{\cdots}}}} \ln a$

Take the natural logarithm once more:

$\ln \left(\ln y\right) = \ln \left({x}^{{a}^{{x}^{{a}^{\cdots}}}} \ln a\right)$

$\textcolor{w h i t e}{\ln} \left(\ln y\right) = \ln \left({x}^{{a}^{{x}^{{a}^{\cdots}}}}\right) + \ln \left(\ln a\right)$

$\textcolor{w h i t e}{\ln} \left(\ln y\right) = {a}^{{x}^{{a}^{{x}^{\cdots}}}} \ln x + \ln \left(\ln a\right)$

$\textcolor{w h i t e}{\ln} \left(\ln y\right) = y \ln x + \ln \left(\ln a\right)$

Taking the derivative of this last version:

$\frac{d}{\mathrm{dx}} \ln \left(\ln y\right) = \frac{d}{\mathrm{dx}} \left(y \ln x\right) + \frac{d}{\mathrm{dx}} \ln \left(\ln a\right)$

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

$\frac{1}{\ln} y \left(\frac{d}{\mathrm{dx}} \ln y\right) = \frac{\mathrm{dy}}{\mathrm{dx}} \ln x + \frac{y}{x} + 0$

$\frac{1}{\ln} y \left(\frac{1}{y}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}} \ln x + \frac{y}{x}$

Grouping $\frac{\mathrm{dy}}{\mathrm{dx}}$ terms:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(\frac{1}{y \ln y} - \ln x\right) = \frac{y}{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(\frac{1 - y \ln x \ln y}{y \ln y}\right) = \frac{y}{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{2} \ln y}{x \left(1 - y \ln x \ln y\right)}$

Sep 9, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{2} L o g y}{x \left(1 - y L o g x L o g y\right)}$

#### Explanation:

If $y = {a}^{{x}^{{a}^{{x}^{\cdots}}}}$

then

$y = {a}^{{x}^{y}}$ then applying $\log$ to both sides

$\log y = {x}^{y} \log \left(a\right)$ and now applying again $\log$ to both sides

$\log \left(\log y\right) = y \log x + \log \left(\log a\right)$ and now deriving

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{2} L o g y}{x \left(1 - y L o g x L o g y\right)}$