Find y' for y = ln [ln(ln x) ]
Find $y'$ for $y = \ln\bigl[\ln(\ln x)\bigr]$.
Explanation of Method
Three nested logarithms, giving three nested factors.
Step 1: Name the layers
\[ w = \ln x, \qquad v = \ln w = \ln(\ln x), \qquad y = \ln v. \]
Step 2: Apply the chain rule
Each logarithm contributes the reciprocal of its own argument: \[y' = \frac{1}{v}\cdot\frac{1}{w}\cdot\frac{1}{x} = \frac{1}{\ln(\ln x)}\cdot\frac{1}{\ln x}\cdot\frac{1}{x}.\]
Step 3: Verification
\[x = e^{e} \approx 15.15426:\quad \ln x = e,\ \ln\ln x = 1 \;\Longrightarrow\; y' = \frac{1}{e^{e}\cdot e\cdot 1} \approx 0.024276\]
\[\frac{y(e^{e}+10^{-4})-y(e^{e}-10^{-4})}{2\times 10^{-4}} \approx 0.024276 \;\;\checkmark\]
Step 4: Domain
Each layer imposes its own condition: $x \gt 0$ for $\ln x$, then $\ln x \gt 0$ for $\ln\ln x$, then $\ln\ln x \gt 0$ for the outermost. The last is the strongest, requiring $\ln x \gt 1$, i.e.
\[ x \gt e \approx 2.71828. \]
Result
Final answer:
\[\boxed{\,y' = \dfrac{1}{x\ln x\cdot\ln(\ln x)}\,}\]
Remark
Remark. The pattern is worth remembering: each extra logarithm multiplies the derivative by the reciprocal of one more nested logarithm. A fourth layer would contribute a further factor $\dfrac{1}{\ln\ln\ln x}$, on the domain $x \gt e^{e}$.