# Thread: stuck on second x^x

1. ## stuck on second x^x

$\displaystyle \begin{gathered} f(x) = {x^{{x^x}}} \hfill \\ \ln y = {x^x}\ln x \hfill \\ \frac{1} {y}y' = \frac{1} {x}*\frac{1} {x} \hfill \\ \end{gathered}$

but for the x^x, ln is needed but is
$\displaystyle \ln \ln y = \ln ({x^x}\ln x)$

okay?

2. Hello, genlovesmusic09!

You are correct . . .

Differentiate: .$\displaystyle y \:=\: x^{x^x}$

Take logs: .$\displaystyle \ln(y) \;=\;\ln\left(x^{x^x}\right) \quad\Rightarrow\quad \ln(y)\;=\;x^x\cdot\ln(x)$

Take logs: .$\displaystyle \ln\left[\ln(y)\right] \;=\;\ln\left[x^x\cdot\ln(x)\right] \;=\; \ln\left[x^x\right] + \ln\left[\ln(x)\right]$

And we have: .$\displaystyle \ln\left[\ln(y)\right] \;=\;x\cdot\ln(x) + \ln\left[\ln(x)\right]$

Differentiate: .$\displaystyle \frac{1}{\ln(y)}\!\cdot\!\frac{1}{y}\!\cdot\! y' \;\;=\;\;x\!\cdot\!\frac{1}{x} \;+\; 1\!\cdot\!\ln(x) \;+\; \frac{1}{\ln(x)}\!\cdot\!\frac{1}{x}$

. . . . . . . . . . . . $\displaystyle \frac{y'}{y\!\cdot\!\ln(y)} \;=\;1 + \ln(x) + \frac{1}{x\!\cdot\!\ln(x)}$

. . . . . . . . . . . . . . . $\displaystyle y' \;=\;y \cdot \ln(y)\cdot\bigg[1 + \ln(x) + \frac{1}{x\!\cdot\!\ln(x)} \bigg]$

. . . . . . . . . . . . . . . $\displaystyle y' \;=\;x^{x^x}\cdot\ln\left(x^{x^x}\right)\cdot\bigg[1 + \ln(x) + \frac{1}{x\!\cdot\!\ln(x)} \bigg]$