求导：$y=(\ln x)^{\ln x}$

取对数：

$$\ln y=\ln x\cdot\ln(\ln x).$$

两边求导：

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

因此

$$\frac{dy}{dx}=(\ln x)^{\ln x}\cdot\frac{\ln(\ln x)+1}{x}.$$

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Original Explanation

Let $$u \;=\;\ln y \;=\;\ln\Bigl((\ln x)^{\,\ln x}\Bigr) \;=\;\ln x \;\times\;\ln(\ln x).$$ Then $$\frac{du}{dx} \;=\; \frac{d(\ln x)}{dx}\,\ln(\ln x) \;+\; \ln x\;\times\;\frac{d(\ln(\ln x))}{dx} \;=\; \frac{\ln(\ln x)}{\,x\,} \;+\; \frac{\ln x}{\,x\,\ln x\,} \;=\; \frac{\ln(\ln x)+1}{\,x\,}.$$ Since $\frac{du}{dx} = \frac{1}{y}\,\frac{dy}{dx},$ we get $$\frac{dy}{dx} \;=\; y \,\frac{du}{dx} \;=\; (\ln x)^{\,\ln x} \;\frac{\ln(\ln x)+1}{\,x\,}.$$