Dziedzina funkcji \(f:\)
\[x\geqslant 0, \ \ x \gt 0, \ \ \ln x\neq 0 \ \Leftrightarrow \ x\neq 1\]
Zatem \(D_{f}=\mathbb{R}_{+}\setminus \left \{ 1 \right \}.\)
\[ \begin{array}{l}
{\displaystyle f(x)=\frac{\sqrt{x}}{\ln x}}\\
{\displaystyle f'(x)=\frac{\frac{1}{2\sqrt{x}}\ln x-\sqrt{x}\cdot \frac{1}{x}}{\ln^{2}x}=
\frac{\frac{\ln x}{2\sqrt{x}}-\frac{1}{\sqrt{x}}}{\ln^{2}x}=
\frac{\frac{\ln x}{2}-1}{\sqrt{x}\ln^{2}x}}\\
{\displaystyle f'(x)\gt 0 \ \Leftrightarrow \ \frac{\frac{\ln x}{2}-1}{\sqrt{x}\ln^{2}x}\gt 0}\\
{\displaystyle \sqrt{x}\ln^{2}x(\frac{\ln x}{2}-1) \gt 0}\\
{\displaystyle \sqrt{x} =0 \ \ \vee \ \ \ln^{2}x =0 \ \ \vee \ \ \frac{\ln x}{2}=1}\\
{\displaystyle x =0 \ \ \vee \ \ \ln x =0 \ \ \vee \ \ \ln x=2}\\
{\displaystyle x =0 \ \ \vee \ \ x =1 \textrm {- p. 2-kr.} \ \ \vee \ \ \ln x=2}\\
{\displaystyle x =0 \ \ \vee \ \ x =1 \textrm {- p. 2-kr.} \ \ \vee \ \ x=e^{2}}\\
\end{array}\]
Zatem uwzględniając dziedzinę \(x\gt 0,\ x\neq 1\) mamy:
\[ \begin{array}{l}
f \searrow \ \Leftrightarrow \ x\in \left ( 0;1 \right )\\
f \searrow \ \Leftrightarrow \ x\in \left ( 1;e^{2} \right )\\
f \nearrow \ \Leftrightarrow \ x\in \left ( e^{2};\infty \right )
\end{array}\]