Lösung 1.2:2d
Aus Online Mathematik Brückenkurs 2
We can see the expression as "ln of something",
where "something" is
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}
