# Thread: stuck on second x^x

1. ## stuck on second x^x

$\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
$\ln \ln y = \ln ({x^x}\ln x)
$

okay?

2. Hello, genlovesmusic09!

You are correct . . .

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

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

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

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

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

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

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

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