Aus Online Mathematik Brückenkurs 2

We can see the expression as "ln of something",

\displaystyle \ln \left\{ \left. {} \right\} \right.,

where "something" is \displaystyle \ln x.

Because we have a compound expression, we use the chain rule and obtain, roughly speaking, the outer derivative multiplied by the inner derivative,

\displaystyle \frac{d}{dx}\ln \left\{ \left. \ln x \right\} \right.=\frac{1}{\ln x}\centerdot \left( \left\{ \left. \ln x \right\} \right. \right)^{\prime }

where the first factor on the right-hand side \displaystyle \frac{1}{\ln x} is the outer derivative of \displaystyle \ln \left\{ \left. \ln x \right\} \right. and the other factor \displaystyle \left( \left\{ \left. \ln x \right\} \right. \right)^{\prime } is the inner derivative. Thus, we get

\displaystyle \frac{d}{dx}\ln \left\{ \left. \ln x \right\} \right.=\frac{1}{\ln x}\centerdot \frac{1}{x}=\frac{1}{x\ln x}