Find y' for y = x^(x^x), x > 0
Find $y'$ for $y = x^{x^{x}}$, $x \gt 0$.
Explanation of Method
A genuine power tower, differentiated one level at a time.
Step 1: Read the tower correctly
By convention $x^{x^{x}}$ means $x^{\left(x^{x}\right)}$, not $\left(x^{x}\right)^{x} = x^{x^{2}}$. Taking logarithms peels off the outermost level: $\ln y = x^{x}\ln x$.
Step 2: Differentiate the right-hand side
This is a product of $x^{x}$ and $\ln x$. Problem 175 supplies $\bigl(x^{x}\bigr)' = x^{x}\bigl(1+\ln x\bigr)$: \[\frac{d}{dx}\bigl(x^{x}\ln x\bigr) = x^{x}\bigl(1+\ln x\bigr)\ln x + x^{x}\cdot\frac1x = x^{x}\left[\ln x + \ln^{2}x + \frac1x\right].\]
Step 3: Solve for $y'$
\[ \frac{y'}{y} = x^{x}\left(\frac1x + \ln x + \ln^{2}x\right) \;\Longrightarrow\; y' = x^{x^{x}}\cdot x^{x}\left(\frac1x + \ln x + \ln^{2}x\right). \]
Step 4: Verification
\[x = 1:\quad y = 1^{1} = 1,\quad y' = 1\cdot 1\cdot(1+0+0) = 1 \;\;\checkmark\]
\[\frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 1.0 \;\;\checkmark\]
\[x = 2:\quad y = 2^{4} = 16,\quad y' = 16\cdot 4\left(\tfrac12 + \ln 2 + \ln^{2}2\right) \approx 107.11 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = x^{x^{x}}\cdot x^{x}\left(\dfrac{1}{x} + \ln x + \ln^{2}x\right), \qquad x > 0\,}\]
Remark
Remark. Each extra level of the tower costs one more logarithm: $x^{x}$ gives $1+\ln x$, and $x^{x^{x}}$ gives $\tfrac1x+\ln x+\ln^{2}x$ together with an extra factor $x^{x}$. The bracket is positive for every $x \gt 0$: substituting $u = \ln x$ turns it into $e^{-u}+u+u^{2}$, whose derivative $-e^{-u}+1+2u$ vanishes only at $u = 0$, where the value is $1$. So the tower is strictly increasing.