Find y' for y = (ln x)^x, x > 1

Find $y'$ for $y = (\ln x)^{x}$, $x \gt 1$.

Step 1: Take the natural logarithm

The exponent $x$ comes down and the base becomes a nested logarithm: $\ln y = x\ln(\ln x)$.

Step 2: Differentiate both sides

Product rule on the right, with $\bigl(\ln\ln x\bigr)' = \dfrac{1}{x\ln x}$ from Problem 124: $\frac{y'}{y} = \ln(\ln x) + x\cdot\frac{1}{x\ln x} = \ln(\ln x) + \frac{1}{\ln x}$.

The factor $x$ cancels against the $\dfrac1x$, leaving no $x$ in the second term.

Step 3: Solve for $y'$

\[ y' = (\ln x)^{x}\left(\ln\ln x + \frac{1}{\ln x}\right). \]

Step 4: Verification

\[x = e:\quad \ln e = 1 \;\Longrightarrow\; y = 1^{e} = 1,\quad y' = 1\cdot\bigl(\ln 1 + 1\bigr) = 1 \;\;\checkmark\]

\[\frac{y(e+10^{-5})-y(e-10^{-5})}{2\times 10^{-5}} \approx 1.0 \;\;\checkmark\]

Step 5: Domain

Two conditions bite. The base must be positive for a non-integer power, so $\ln x \gt 0$, i.e. $x \gt 1$; and the term $\ln\ln x$ in the derivative needs the same.

Although $\ln\ln x \lt 0$ for $1 \lt x \lt e$, the bracket never vanishes. Substituting $u = \ln x \gt 0$ turns it into $h(u) = \ln u + \dfrac1u$, and

\[ h'(u) = \frac{1}{u} - \frac{1}{u^{2}} = \frac{u-1}{u^{2}}, \qquad h(1) = 0 + 1 = 1. \]

So $h$ has its minimum at $u = 1$ (that is, $x = e$) with value $1 \gt 0$. Hence $y' \gt 0$ throughout and $y = (\ln x)^{x}$ is strictly increasing on $(1,\infty)$.